A rock is thrown directly upward from the edge of the roof of a building that is 34.6m tall (kinematics)?
Q. A rock is thrown directly upward from the edge of the roof of a building that is 34.6m tall. The rock misses the building on its way down and is observed to strike the ground 4.0 seconds after being thrown. With what speed was the rock thrown? Can someone give me a detailed explanation please?
Asked by Kevin S - Wed Oct 21 12:58:42 2009 - - 2 Answers - 0 Comments
A. use the equation of motion: y(t)=y0+v0t-1/2gt^2 y(t)=height of projectile at any time, t y0=initial height = 34.6 m v0=initial speed (to be determined) t=time of flight we are told that the ball hits the ground (y=0) at t=4, so we have 0=34.6+v0(4s)-1/2 (9.8m/s/s)(4s)^2 0=34.6+4v0-78.4 4v0=43.8 v0=10.95m/s
Answered by kuiperbelt2003 - Wed Oct 21 13:04:41 2009
Q. A rock is thrown directly upward from the edge of the roof of a building that is 34.6m tall. The rock misses the building on its way down and is observed to strike the ground 4.0 seconds after being thrown. With what speed was the rock thrown? Can someone give me a detailed explanation please?
Asked by Kevin S - Wed Oct 21 12:58:42 2009 - - 2 Answers - 0 Comments
A. use the equation of motion: y(t)=y0+v0t-1/2gt^2 y(t)=height of projectile at any time, t y0=initial height = 34.6 m v0=initial speed (to be determined) t=time of flight we are told that the ball hits the ground (y=0) at t=4, so we have 0=34.6+v0(4s)-1/2 (9.8m/s/s)(4s)^2 0=34.6+4v0-78.4 4v0=43.8 v0=10.95m/s
Answered by kuiperbelt2003 - Wed Oct 21 13:04:41 2009
Rock thrown from building - free fall physics question ?
Q. A rock is thrown vertically upward with a speed of 19.0 m/s from the roof of a building that is 70.0 m above the ground. Assume free fall. In how many seconds after being thrown does the rock strike the ground? What is the speed of the rock just before it strikes the ground? p.s. thank you in advance!
Asked by Christine R - Thu Jan 22 23:26:26 2009 - - 1 Answers - 0 Comments
A. Vi = 19 d = -70 a = -9.8 Although once thrown upward, the distance will be greater than 70m, however the Displacement is only 70. d = Vi*t + 1/2at^2 1/2at^2 + Vi*t - d = 0 (Solve the quadratic equation for t) -4.9t^2 + 19t + 70 = 0 -19 (19^2 -4(-4.9)(70) / 2(-4.9) = t t = -19 41.63 / -9.8 t = -2.31, 6.2 Since time cannot be negative, the answer is 6.2 seconds. Vf = Vi + at Vf = -41.76 m/s
Answered by Twixter - Fri Jan 23 00:00:36 2009
Q. A rock is thrown vertically upward with a speed of 19.0 m/s from the roof of a building that is 70.0 m above the ground. Assume free fall. In how many seconds after being thrown does the rock strike the ground? What is the speed of the rock just before it strikes the ground? p.s. thank you in advance!
Asked by Christine R - Thu Jan 22 23:26:26 2009 - - 1 Answers - 0 Comments
A. Vi = 19 d = -70 a = -9.8 Although once thrown upward, the distance will be greater than 70m, however the Displacement is only 70. d = Vi*t + 1/2at^2 1/2at^2 + Vi*t - d = 0 (Solve the quadratic equation for t) -4.9t^2 + 19t + 70 = 0 -19 (19^2 -4(-4.9)(70) / 2(-4.9) = t t = -19 41.63 / -9.8 t = -2.31, 6.2 Since time cannot be negative, the answer is 6.2 seconds. Vf = Vi + at Vf = -41.76 m/s
Answered by Twixter - Fri Jan 23 00:00:36 2009
A rock is thrown vertically upward with a speed of 16.0 from the roof of a building that is 70.0 above...?
Q. A rock is thrown vertically upward with a speed of 16.0 from the roof of a building that is 70.0 above the ground. Assume free fall. A rock is thrown vertically upward with a speed of 16.0 m/s from the roof of a building that is 70.0 m above the ground. Assume free fall. How long does it take to hit the ground? At what speed will it hit the ground?
Asked by squakajaneenayoc - Wed Oct 7 00:01:59 2009 - - 1 Answers - 0 Comments
Q. A rock is thrown vertically upward with a speed of 16.0 from the roof of a building that is 70.0 above the ground. Assume free fall. A rock is thrown vertically upward with a speed of 16.0 m/s from the roof of a building that is 70.0 m above the ground. Assume free fall. How long does it take to hit the ground? At what speed will it hit the ground?
Asked by squakajaneenayoc - Wed Oct 7 00:01:59 2009 - - 1 Answers - 0 Comments
A rock is thrown upward with speed of 21.0m/s from the roof of a building that is 42.0 m above ground. Find t
Q. In how many seconds after being thrown does the rock strike the ground? What is the speed of the rock just before it strikes the ground? Thanks in adavance.
Asked by bouncing_soul_geuse - Thu Sep 13 06:09:52 2007 - - 2 Answers - 0 Comments
A. ok. x= 0 m xo= 42 m v= ? vo= 21 m/s a= -9.8 m/s/s t = ? solve for v with this equation- v^2= vo+.5a(x-xo) to solve for time use this equation- v= vo+at you should get a negative velocity in the end. and a time of about 5 seconds. velocity should be around -30 m/s. answer 2 got it right. except the velocty is negative because its coming down.
Answered by Reds4Life - Thu Sep 13 06:24:12 2007
Q. In how many seconds after being thrown does the rock strike the ground? What is the speed of the rock just before it strikes the ground? Thanks in adavance.
Asked by bouncing_soul_geuse - Thu Sep 13 06:09:52 2007 - - 2 Answers - 0 Comments
A. ok. x= 0 m xo= 42 m v= ? vo= 21 m/s a= -9.8 m/s/s t = ? solve for v with this equation- v^2= vo+.5a(x-xo) to solve for time use this equation- v= vo+at you should get a negative velocity in the end. and a time of about 5 seconds. velocity should be around -30 m/s. answer 2 got it right. except the velocty is negative because its coming down.
Answered by Reds4Life - Thu Sep 13 06:24:12 2007
physics problem...A rock is thrown vertically upward...?
Q. A rock is thrown vertically upward with a speed of 18.0m/s from the roof of a building that is 70.0m above the ground. Assume free fall. In how many seconds after being thrown does the rock strike the ground? What is the speed of the rock just before it strikes the ground?
Asked by ! - Thu Sep 6 22:36:09 2007 - - 1 Answers - 0 Comments
A. First find the max height of the rock: ymax = (V^2 - V0^2)/(2g) where v0 = 18 m/s V = 0 and g = -9.8 m/s^2 ymax = 18^2/(2*9.8) = 16.53 m from top of building. Now rock falls 16.53 + 70 = 86.53 m and starts with V0 = 0 so y = -86.53 = V^2/(2*g) ---> V =sqrt(2*9.8*86.53) = -41.18 m/s Now we can compute time: Time to go to max height is found using v = gt+v0 = 0 =-9.8 t + 18 ---> t = 18/9.8 = 1.84 sec Now time to go from max height to ground is: v =gt = -41.18 = -9.8 t ---> t =41.18/9.8 = 4.2 sec Total time = 4.2 + 1.84 = 6.04 sec
Answered by nyphdinmd - Thu Sep 6 22:48:49 2007
Q. A rock is thrown vertically upward with a speed of 18.0m/s from the roof of a building that is 70.0m above the ground. Assume free fall. In how many seconds after being thrown does the rock strike the ground? What is the speed of the rock just before it strikes the ground?
Asked by ! - Thu Sep 6 22:36:09 2007 - - 1 Answers - 0 Comments
A. First find the max height of the rock: ymax = (V^2 - V0^2)/(2g) where v0 = 18 m/s V = 0 and g = -9.8 m/s^2 ymax = 18^2/(2*9.8) = 16.53 m from top of building. Now rock falls 16.53 + 70 = 86.53 m and starts with V0 = 0 so y = -86.53 = V^2/(2*g) ---> V =sqrt(2*9.8*86.53) = -41.18 m/s Now we can compute time: Time to go to max height is found using v = gt+v0 = 0 =-9.8 t + 18 ---> t = 18/9.8 = 1.84 sec Now time to go from max height to ground is: v =gt = -41.18 = -9.8 t ---> t =41.18/9.8 = 4.2 sec Total time = 4.2 + 1.84 = 6.04 sec
Answered by nyphdinmd - Thu Sep 6 22:48:49 2007
How do I solve this physics problem?
Q. A man stands on the roof of a 13.0 m-tall building and throws a rock with a velocity of magnitude 30.0 m at an angle of 42.0 above the horizontal. You can ignore air resistance. Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. I keep getting 58.4 but it keeps saying im wrong, please help! Thanks guys!
Asked by Psiboi - Wed Sep 24 15:11:57 2008 - - 2 Answers - 0 Comments
A. The upward-speed of the ball is 30 * sin(42 degrees) = 20.074m/s The motion equation is delta H = Vh(start) * deltaT + 0.5 * (-g) * deltaT^2 we know deltaH = (-13m) (the height of the building) therefor -13 = 20.074*dT - 4.9dT^2 solving it we get dT = 4.665 seconds the Horizontal speed of the rock is 30 * cos(42 deg) = 22.294 m/s therefor, assuming constant velocity horizontally, we get Distance = 22.294 m/s * 4.665 seconds = 104 m
Answered by Barak F - Wed Sep 24 15:32:50 2008
Q. A man stands on the roof of a 13.0 m-tall building and throws a rock with a velocity of magnitude 30.0 m at an angle of 42.0 above the horizontal. You can ignore air resistance. Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. I keep getting 58.4 but it keeps saying im wrong, please help! Thanks guys!
Asked by Psiboi - Wed Sep 24 15:11:57 2008 - - 2 Answers - 0 Comments
A. The upward-speed of the ball is 30 * sin(42 degrees) = 20.074m/s The motion equation is delta H = Vh(start) * deltaT + 0.5 * (-g) * deltaT^2 we know deltaH = (-13m) (the height of the building) therefor -13 = 20.074*dT - 4.9dT^2 solving it we get dT = 4.665 seconds the Horizontal speed of the rock is 30 * cos(42 deg) = 22.294 m/s therefor, assuming constant velocity horizontally, we get Distance = 22.294 m/s * 4.665 seconds = 104 m
Answered by Barak F - Wed Sep 24 15:32:50 2008
what are the answers to the following problems?
Q. A diver running 1.0 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.2 s later. How high was the cliff and how far from its base did the diver hit the water? A ball is thrown horizontally from the roof of a building 56 m tall and lands 40 m from the base. What was the ball's initial speed? A tiger leaps horizontally from a 6.0 m high rock with a speed of 7.0 m/s. How far from the base of the rock will she land? A ball thrown horizontally at 22.1 m/s from the roof of a building lands 36.0 m from the base of the building. How high is the building? A football is kicked at ground level with a speed of 20.0 m/s at an angle of 39.0 to the horizontal. How much later does it hit the ground?
Asked by ggggg - Thu Oct 11 21:03:41 2007 - - 1 Answers - 0 Comments
A. Are you trying to get people to do your homework? The answer is: get a smart kid to do it for you.
Answered by Sara - Thu Oct 11 21:11:52 2007
Q. A diver running 1.0 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.2 s later. How high was the cliff and how far from its base did the diver hit the water? A ball is thrown horizontally from the roof of a building 56 m tall and lands 40 m from the base. What was the ball's initial speed? A tiger leaps horizontally from a 6.0 m high rock with a speed of 7.0 m/s. How far from the base of the rock will she land? A ball thrown horizontally at 22.1 m/s from the roof of a building lands 36.0 m from the base of the building. How high is the building? A football is kicked at ground level with a speed of 20.0 m/s at an angle of 39.0 to the horizontal. How much later does it hit the ground?
Asked by ggggg - Thu Oct 11 21:03:41 2007 - - 1 Answers - 0 Comments
A. Are you trying to get people to do your homework? The answer is: get a smart kid to do it for you.
Answered by Sara - Thu Oct 11 21:11:52 2007
physics help (free falling object)?
Q. A rock is thrown directly upward from the edge of the roof of a building that is 31.3 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Take the acceleration due to gravity to have magnitude 9.80m/s^2 and neglect any effects of air resistance. With what speed was the rock thrown?
Asked by RisingPhoenix - Sat Feb 7 00:44:49 2009 - - 2 Answers - 0 Comments
A. S=ut + 1/2at2 (for the whole motion) upwards, -31.3m =u*4s + 1/2*(-)9.80m/s2 *4s*4s -31.3 = 4u - 78.4 -31.3 + 78.4 = 4u 56.1 = 4u u = 56.1/4 u = 14.025 m/s Speed = 14.025 m/s Hope the above is clear!
Answered by i_need_answers - Sat Feb 7 04:09:33 2009
Q. A rock is thrown directly upward from the edge of the roof of a building that is 31.3 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Take the acceleration due to gravity to have magnitude 9.80m/s^2 and neglect any effects of air resistance. With what speed was the rock thrown?
Asked by RisingPhoenix - Sat Feb 7 00:44:49 2009 - - 2 Answers - 0 Comments
A. S=ut + 1/2at2 (for the whole motion) upwards, -31.3m =u*4s + 1/2*(-)9.80m/s2 *4s*4s -31.3 = 4u - 78.4 -31.3 + 78.4 = 4u 56.1 = 4u u = 56.1/4 u = 14.025 m/s Speed = 14.025 m/s Hope the above is clear!
Answered by i_need_answers - Sat Feb 7 04:09:33 2009
Projectile motion help please?Thank you very much!?
Q. 1)A girl standing on the top of a roof throws a rock at 30m/s at an angle of 30 below the horizontal. If the roof is 50m high, how far from the base of the building will the rock land.?
Asked by Eric F - Wed Mar 11 03:28:17 2009 - - 3 Answers - 0 Comments
A. x = 30m/s x cos(30) = 25.98m/s y = 30m/s x sin(30) = 15m/s v^2 = vo^2 + 2ay v^2 = (-15m/s)^2 + 2(-9.8m/s^2)(-50m) v = -34.71m/s v = vo + at -34.71m/s = -15m/s + (-9.8m/s^2)t t = 2.011s x = vt x = (25.98m/s)(2.011s) x = 52.2457m or rounded down to two sig figs = 52m from base
Answered by jj l - Wed Mar 11 03:42:51 2009
Q. 1)A girl standing on the top of a roof throws a rock at 30m/s at an angle of 30 below the horizontal. If the roof is 50m high, how far from the base of the building will the rock land.?
Asked by Eric F - Wed Mar 11 03:28:17 2009 - - 3 Answers - 0 Comments
A. x = 30m/s x cos(30) = 25.98m/s y = 30m/s x sin(30) = 15m/s v^2 = vo^2 + 2ay v^2 = (-15m/s)^2 + 2(-9.8m/s^2)(-50m) v = -34.71m/s v = vo + at -34.71m/s = -15m/s + (-9.8m/s^2)t t = 2.011s x = vt x = (25.98m/s)(2.011s) x = 52.2457m or rounded down to two sig figs = 52m from base
Answered by jj l - Wed Mar 11 03:42:51 2009
Physics problem!! pls help.. tnx?
Q. A man stands on the roof of a 15 m tall building and throws a rock with a velocity of magnitude 30 m/s at an angle 33 degrees above the horizontal. You can ignore air resistance. calculate a) the maximum height above the roof reached by the rock. b) the magnitude of the velocity of the rock just before it strikes the ground. c) the horizontal distance from the base of the building to the point where the rock strikes the ground.
Asked by Darlane - Tue Sep 1 13:13:00 2009 - - 2 Answers - 0 Comments
A. Ok here we go...!!! - angle of projection...in question - 33o a) - We are given that the initial velocity, u = 30m/s... = 33o The max.Height reached my the projectile, H = u^2sin^2 /2g H = 30 x 30 x 0.54 x 0.54/ 2 x 9.8 H = 13.38m...therefore the stone reaches a height of 13.38 m above the roof...(ANSWER) b) - Inorder to calculate velocity with which the stone strikes the ground...we have to consider the x-axis component of final velocity as Vx and y-axis component of final velocity as Vy... now by newton;s laws of motion...Vx = ux + ax x t given that the initial velocity, u = 30m/s...therefore x - axis component of initial velocity, ux = usin = 30 x 0.54 = 16.2m/s Also ax is always equal to zero because acceleration acts… [cont.]
Answered by unknown - Tue Sep 1 14:05:46 2009
Q. A man stands on the roof of a 15 m tall building and throws a rock with a velocity of magnitude 30 m/s at an angle 33 degrees above the horizontal. You can ignore air resistance. calculate a) the maximum height above the roof reached by the rock. b) the magnitude of the velocity of the rock just before it strikes the ground. c) the horizontal distance from the base of the building to the point where the rock strikes the ground.
Asked by Darlane - Tue Sep 1 13:13:00 2009 - - 2 Answers - 0 Comments
A. Ok here we go...!!! - angle of projection...in question - 33o a) - We are given that the initial velocity, u = 30m/s... = 33o The max.Height reached my the projectile, H = u^2sin^2 /2g H = 30 x 30 x 0.54 x 0.54/ 2 x 9.8 H = 13.38m...therefore the stone reaches a height of 13.38 m above the roof...(ANSWER) b) - Inorder to calculate velocity with which the stone strikes the ground...we have to consider the x-axis component of final velocity as Vx and y-axis component of final velocity as Vy... now by newton;s laws of motion...Vx = ux + ax x t given that the initial velocity, u = 30m/s...therefore x - axis component of initial velocity, ux = usin = 30 x 0.54 = 16.2m/s Also ax is always equal to zero because acceleration acts… [cont.]
Answered by unknown - Tue Sep 1 14:05:46 2009
best answer would be rewarded.. pls answer the problems about physics?
Q. 1. A man stands on the roof of a 15 m. tall building and throws a rock with a velocity of magnitude 30 m/s at an angle of 33 degrees above the horizontal. You can ignore the air resistance. Calculate. a. the maximum height above the roof reached by the rock; b. magnitude of the velocity of the rock just before it strikes the ground; c. the horizontal distance from the base of the building to the point where the rock strikes the ground. 2. A snowball rolls off a barn roof that slopes downward at an angle of 40 degrees. The edge of the roof is 14 m. above the ground, and the snowball has a speed of 7 m/s as it rolls off the roof. Ignore air resistance. a. How far from the edge of the barn does the snowball strike the ground if it doesn t… [cont.]
Asked by eden - Fri Sep 11 07:27:16 2009 - - 1 Answers - 0 Comments
A. maximum height= u^2sin^2(Theta)/2g
Answered by Manu - Fri Sep 11 09:03:53 2009
Q. 1. A man stands on the roof of a 15 m. tall building and throws a rock with a velocity of magnitude 30 m/s at an angle of 33 degrees above the horizontal. You can ignore the air resistance. Calculate. a. the maximum height above the roof reached by the rock; b. magnitude of the velocity of the rock just before it strikes the ground; c. the horizontal distance from the base of the building to the point where the rock strikes the ground. 2. A snowball rolls off a barn roof that slopes downward at an angle of 40 degrees. The edge of the roof is 14 m. above the ground, and the snowball has a speed of 7 m/s as it rolls off the roof. Ignore air resistance. a. How far from the edge of the barn does the snowball strike the ground if it doesn t… [cont.]
Asked by eden - Fri Sep 11 07:27:16 2009 - - 1 Answers - 0 Comments
A. maximum height= u^2sin^2(Theta)/2g
Answered by Manu - Fri Sep 11 09:03:53 2009
rate of change problem?
Q. A rock is thrown down off the roof of a building with an initial speed of 9.8m/s and hits the ground at the speed of 49m/s. how high is the building, a=9.8 m/s^2
Asked by Shinu - Tue Nov 13 20:40:27 2007 - - 2 Answers - 0 Comments
A. There are four variables which put together in an equation can describe this motion. These are Initial Velocity (u); Final Velocity (v), Acceleration (a), Distance Traveled (s) and Time elapsed (t). The equations which tell us the relationship between these variables are as given below. v = u + at v2 = u2 + 2as s = ut + 1/2 at2 49^2 = 9.8^2 + 2(9.8)s solve this and youll get the answer
Answered by Jay - Tue Nov 13 20:46:13 2007
Q. A rock is thrown down off the roof of a building with an initial speed of 9.8m/s and hits the ground at the speed of 49m/s. how high is the building, a=9.8 m/s^2
Asked by Shinu - Tue Nov 13 20:40:27 2007 - - 2 Answers - 0 Comments
A. There are four variables which put together in an equation can describe this motion. These are Initial Velocity (u); Final Velocity (v), Acceleration (a), Distance Traveled (s) and Time elapsed (t). The equations which tell us the relationship between these variables are as given below. v = u + at v2 = u2 + 2as s = ut + 1/2 at2 49^2 = 9.8^2 + 2(9.8)s solve this and youll get the answer
Answered by Jay - Tue Nov 13 20:46:13 2007
physics helps please projectile motion?
Q. A man stands on the roof of a 20.0 m tall building and throws a rock with a velocity of magnitude 20.0 m/s at an angle of 30.0 above the horizontal. You can ignore air resistance. (a) Calculate the maximum height above the roof reached by the rock. (b) Calculate the magnitude of the velocity of the rock just before it strikes the ground. (c) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. I got the first part but for some stupid reason i cant get the other two. Thanks so much!
Asked by Hannah S - Sun May 18 15:41:55 2008 - - 1 Answers - 0 Comments
A. (a) 1/2m(vsin30)^2=mgh or h=(vsin30)^2/(2*g) (b) for vertical axis: 1/2*m(v1)^2=mg(h+20) horizontal: v2=vcos30 v=sqr(v1^2+v2^2) (c) free fall from h: h=1/2gt^2 solve for t horizontal motion: x=vcos30*t
Answered by Dimitris P - Sun May 18 16:33:39 2008
Q. A man stands on the roof of a 20.0 m tall building and throws a rock with a velocity of magnitude 20.0 m/s at an angle of 30.0 above the horizontal. You can ignore air resistance. (a) Calculate the maximum height above the roof reached by the rock. (b) Calculate the magnitude of the velocity of the rock just before it strikes the ground. (c) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. I got the first part but for some stupid reason i cant get the other two. Thanks so much!
Asked by Hannah S - Sun May 18 15:41:55 2008 - - 1 Answers - 0 Comments
A. (a) 1/2m(vsin30)^2=mgh or h=(vsin30)^2/(2*g) (b) for vertical axis: 1/2*m(v1)^2=mg(h+20) horizontal: v2=vcos30 v=sqr(v1^2+v2^2) (c) free fall from h: h=1/2gt^2 solve for t horizontal motion: x=vcos30*t
Answered by Dimitris P - Sun May 18 16:33:39 2008
What is the angular momentum as a function of time?
Q. A rock of mass 60 g is thrown with initial horizontal speed Vx=25m/s off a building from height of 30m. Calculate the angular momentum of the rock about the line along the edge of the roof as a function of time.
Asked by Dancer - Sun Aug 16 16:19:53 2009 - - 1 Answers - 0 Comments
A. You like dancing, right?
Answered by Happy Face - Sun Aug 16 17:12:22 2009
Q. A rock of mass 60 g is thrown with initial horizontal speed Vx=25m/s off a building from height of 30m. Calculate the angular momentum of the rock about the line along the edge of the roof as a function of time.
Asked by Dancer - Sun Aug 16 16:19:53 2009 - - 1 Answers - 0 Comments
A. You like dancing, right?
Answered by Happy Face - Sun Aug 16 17:12:22 2009
Free fall problem anyone?
Q. A man stands on the roof of a building of height 13.0 m and throws a rock with a velocity of magnitude 25.5 m/s at an angle of 32.1 degrees above the horizontal. You can ignore air resistance. Calculate the maximum height above the roof reached by the rock. Take free fall acceleration to be 9.80 m/s^2 Calculate the magnitude of the velocity of the rock just before it strikes the ground. Take free fall acceleration to be 9.80 m/s^2 Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. Take free fall acceleration to be 9.80 m/s^2
Asked by GUVNA? - Thu Jun 29 01:31:46 2006 - - 4 Answers - 0 Comments
A. Ok, lets see here... 1st step is to break velocity into horizontal and vertical components (think right triangle here): vertical velocity = 25.5 * sin(32.1) = 13.55 m/s horizontal velocity = 25.5 * cos(32.1) = 21.60 m/s Part 1: Figure out how many seconds before the rock's vertical velocity gets to zero (peak of arc): time up = 13.55 / 9.8 = 1.383 sec Find the average vertical velocity during that time with linear acceleration from 13.55 m/s to zero: Vavg up = 13.55 / 2 = 6.775 m/s Find the distance based on average speed and time: distance up = 6.775 * 1.383 = 9.368 meters above the roof Part 2: So we know that the rock is at 22.368 meters above the ground with zero vertical velocity. Now, figure out what it's vertical velocity… [cont.]
Answered by tom_2727 - Thu Jun 29 22:11:03 2006
Q. A man stands on the roof of a building of height 13.0 m and throws a rock with a velocity of magnitude 25.5 m/s at an angle of 32.1 degrees above the horizontal. You can ignore air resistance. Calculate the maximum height above the roof reached by the rock. Take free fall acceleration to be 9.80 m/s^2 Calculate the magnitude of the velocity of the rock just before it strikes the ground. Take free fall acceleration to be 9.80 m/s^2 Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. Take free fall acceleration to be 9.80 m/s^2
Asked by GUVNA? - Thu Jun 29 01:31:46 2006 - - 4 Answers - 0 Comments
A. Ok, lets see here... 1st step is to break velocity into horizontal and vertical components (think right triangle here): vertical velocity = 25.5 * sin(32.1) = 13.55 m/s horizontal velocity = 25.5 * cos(32.1) = 21.60 m/s Part 1: Figure out how many seconds before the rock's vertical velocity gets to zero (peak of arc): time up = 13.55 / 9.8 = 1.383 sec Find the average vertical velocity during that time with linear acceleration from 13.55 m/s to zero: Vavg up = 13.55 / 2 = 6.775 m/s Find the distance based on average speed and time: distance up = 6.775 * 1.383 = 9.368 meters above the roof Part 2: So we know that the rock is at 22.368 meters above the ground with zero vertical velocity. Now, figure out what it's vertical velocity… [cont.]
Answered by tom_2727 - Thu Jun 29 22:11:03 2006
Projectile Motion?
Q. A man stands on the roof of a building of height 16.6 m and throws a rock with a velocity of magnitude 31.1 m/s at an angle of 27.3' above the horizontal. You can ignore air resistance. *The maximum height above the roof reached by the rock: 10.4 m *Take free fall acceleration to be 9.80 m/s^2 A) Calculate the magnitude of the velocity of the rock just before it strikes the ground. B) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. **please show how you got your answer!! THNX!!!**
Asked by Dana - Wed Oct 18 18:11:25 2006 - - 1 Answers - 0 Comments
A. A) You can get the final vertical velocity in one step or two. One step: V^2 = Vo^2 + 2*a*y where V is the final velocity, Vo=0 (at the top of the trajectory), and y = the height it fell from - the top of the trajectory. Two steps: [Might as well use this because it'll give you time and you'll need time to solve B).] y = (1/2)*g*t^2 where y is the same as above and you were given g. Then take that t and use it in V = Vo + g*t where Vo and g are as above. The V you get from that is only the final vertical velocity. Add the horizontal velocity, 31.1*cos(angle) m/s. (Form the resultant vector.) B) The horizontal velocity while all the vertical stuff is going on is 31.1*cos(angle) m/s. Multiply by the time.
Answered by sojsail - Wed Oct 18 19:31:35 2006
Q. A man stands on the roof of a building of height 16.6 m and throws a rock with a velocity of magnitude 31.1 m/s at an angle of 27.3' above the horizontal. You can ignore air resistance. *The maximum height above the roof reached by the rock: 10.4 m *Take free fall acceleration to be 9.80 m/s^2 A) Calculate the magnitude of the velocity of the rock just before it strikes the ground. B) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. **please show how you got your answer!! THNX!!!**
Asked by Dana - Wed Oct 18 18:11:25 2006 - - 1 Answers - 0 Comments
A. A) You can get the final vertical velocity in one step or two. One step: V^2 = Vo^2 + 2*a*y where V is the final velocity, Vo=0 (at the top of the trajectory), and y = the height it fell from - the top of the trajectory. Two steps: [Might as well use this because it'll give you time and you'll need time to solve B).] y = (1/2)*g*t^2 where y is the same as above and you were given g. Then take that t and use it in V = Vo + g*t where Vo and g are as above. The V you get from that is only the final vertical velocity. Add the horizontal velocity, 31.1*cos(angle) m/s. (Form the resultant vector.) B) The horizontal velocity while all the vertical stuff is going on is 31.1*cos(angle) m/s. Multiply by the time.
Answered by sojsail - Wed Oct 18 19:31:35 2006
Physics help please!!?
Q. A rock is thrown vertically down from the roof a 25 meter high building with a speed of 5 m/s. a. when does the rock hit the ground? b. with what speed does it hit the ground? I'm confused on which equation to use. If any of you could just provide some insight onto which equation should be used to solve this, pleaase tell! thanks bunch!!
Asked by Andy - Thu Oct 4 23:58:12 2007 - - 3 Answers - 0 Comments
A. a. use d=v1t+1/2*gt^2 b. use v2=v1+gt
Answered by zsm28 - Fri Oct 5 00:20:23 2007
Q. A rock is thrown vertically down from the roof a 25 meter high building with a speed of 5 m/s. a. when does the rock hit the ground? b. with what speed does it hit the ground? I'm confused on which equation to use. If any of you could just provide some insight onto which equation should be used to solve this, pleaase tell! thanks bunch!!
Asked by Andy - Thu Oct 4 23:58:12 2007 - - 3 Answers - 0 Comments
A. a. use d=v1t+1/2*gt^2 b. use v2=v1+gt
Answered by zsm28 - Fri Oct 5 00:20:23 2007
physics help kinematics?
Q. A rock is thrown vertically down from the roof of a 25.0 m high building with a speed of 5.0 ms ^-1 a) determine the rocks impact velocity, that is, its velocity just as it hits the ground. b) determine the time taken for the rock to reach the ground.
Asked by Paige M - Thu Sep 27 12:44:41 2007 - - 3 Answers - 0 Comments
A. mg h = 0.5m(vf^2 - vi^2) mass on both sides so we get g h = 0.5(vf^2 - vi^2) (9.81)(25) = (0.5)(Vf^2 - 0.2) Vf^2 = 490.7 Vf = 22.15 Once you have Vf finding time is fairly easy.
Answered by aero375 - Thu Sep 27 13:08:15 2007
Q. A rock is thrown vertically down from the roof of a 25.0 m high building with a speed of 5.0 ms ^-1 a) determine the rocks impact velocity, that is, its velocity just as it hits the ground. b) determine the time taken for the rock to reach the ground.
Asked by Paige M - Thu Sep 27 12:44:41 2007 - - 3 Answers - 0 Comments
A. mg h = 0.5m(vf^2 - vi^2) mass on both sides so we get g h = 0.5(vf^2 - vi^2) (9.81)(25) = (0.5)(Vf^2 - 0.2) Vf^2 = 490.7 Vf = 22.15 Once you have Vf finding time is fairly easy.
Answered by aero375 - Thu Sep 27 13:08:15 2007
Physics Help Please explain?
Q. A rock is thrown straight up with an initial speed of 8.00 m/s from the roof of a building 12.0m above the ground. For the motion from the roof to the ground, what are the magnitude and the direction of: (a) the average velocity of the rock? (b) the average acceleration of the rock?
Asked by Imran - Mon Jul 23 23:09:11 2007 - - 4 Answers - 0 Comments
A. a) The average velocity only depends on the initial and final positions of the rock (that is, the roof and the ground). Find the total time of the flight: You know: v0 (= 8.0 m/s), d (= -12.0 m), a (= -9.8 m/s ) You want: t So use this equation: d = v0 t + (1/2) a t (1/2)a t + v0 t - d = 0 t = [-v0 (v0 + 2a d)] / a = [-8 (64 + 235.2)] / -9.8 = 2.58 s v(avg) = d / t = -8 m / 2.58 s = -3.1 m/s b) a(avg) = -g = -9.8 m/s
Answered by Blue - Mon Jul 23 23:23:31 2007
Q. A rock is thrown straight up with an initial speed of 8.00 m/s from the roof of a building 12.0m above the ground. For the motion from the roof to the ground, what are the magnitude and the direction of: (a) the average velocity of the rock? (b) the average acceleration of the rock?
Asked by Imran - Mon Jul 23 23:09:11 2007 - - 4 Answers - 0 Comments
A. a) The average velocity only depends on the initial and final positions of the rock (that is, the roof and the ground). Find the total time of the flight: You know: v0 (= 8.0 m/s), d (= -12.0 m), a (= -9.8 m/s ) You want: t So use this equation: d = v0 t + (1/2) a t (1/2)a t + v0 t - d = 0 t = [-v0 (v0 + 2a d)] / a = [-8 (64 + 235.2)] / -9.8 = 2.58 s v(avg) = d / t = -8 m / 2.58 s = -3.1 m/s b) a(avg) = -g = -9.8 m/s
Answered by Blue - Mon Jul 23 23:23:31 2007
Newton's Laws vs. Energy conservation?
Q. So I'm throwing a rock off a roof with height "H" at a speed "v." I throw a rock 3 times, once straight up, once straight down, and once horizontally away from the building. If I want to find the speed of the rock at which the rock hits the ground fastest so I can decide which is fastest when it hits, should I use Newton's Laws or energy conservation for each of the three cases? I'm still confused with the conceptual part to this problem, and was wondering if anyone could help out. Thanks a lot! =]
Asked by chemistry_nerd - Sun May 3 04:15:30 2009 - - 2 Answers - 0 Comments
A. In either case, you likely need to know the initial velocity of your throw - how do you propose to measure it? I would favour using constant acceleration formulae for this problem - energy conservation principles are faster but constant acceleration formulae can be chosen for the specific problem (i.e. solve for final velocity, or time, or distance etc.) Take for example the case where you throw the rock straight up. It will obey the constant acceleration equation: s = ut + at (s is the position in metres; u is the initial vertical velocity in m/s; a is the acceleration due to gravity (-9.81m/s ); t is time taken for the rock to hit the ground in seconds) So provided you know both the initial speed, OR you can measure how high you… [cont.]
Answered by Oms at the Proms [Deletion] - Sun May 3 05:38:41 2009
Q. So I'm throwing a rock off a roof with height "H" at a speed "v." I throw a rock 3 times, once straight up, once straight down, and once horizontally away from the building. If I want to find the speed of the rock at which the rock hits the ground fastest so I can decide which is fastest when it hits, should I use Newton's Laws or energy conservation for each of the three cases? I'm still confused with the conceptual part to this problem, and was wondering if anyone could help out. Thanks a lot! =]
Asked by chemistry_nerd - Sun May 3 04:15:30 2009 - - 2 Answers - 0 Comments
A. In either case, you likely need to know the initial velocity of your throw - how do you propose to measure it? I would favour using constant acceleration formulae for this problem - energy conservation principles are faster but constant acceleration formulae can be chosen for the specific problem (i.e. solve for final velocity, or time, or distance etc.) Take for example the case where you throw the rock straight up. It will obey the constant acceleration equation: s = ut + at (s is the position in metres; u is the initial vertical velocity in m/s; a is the acceleration due to gravity (-9.81m/s ); t is time taken for the rock to hit the ground in seconds) So provided you know both the initial speed, OR you can measure how high you… [cont.]
Answered by Oms at the Proms [Deletion] - Sun May 3 05:38:41 2009
From Yahoo Answer Search: 'a rock is thrown from the roof of a building'
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One of my favorites is the paleolithic visual symphony known as "Newspaper . Rock. ," an explosion of collective pictography located in Canyonlands National Park in Utah. [Figure 1] [Figure 1: Newspaper . Rock. , Canyonlands National Park, . ... [ Figure 10] [Figure 10: Unidentified photographer, Charles Schreyvogel Painting on the . Roof. of his Apartment . Building. in Hoboken New Jersey, 1903] When this painting is compared to two made by artists who had personal experience of what ...
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