A ball is thrown upward at 20 m/s from the top of a building, which is 100 meter

Question

A ball is thrown upward at 20 m/s from the top of a building, which is 100 meter high. Another ball is thrown downward from the top of the building at 15 m/s. There is a window on the building, which is 2.5 meters high and the top of the window is 60 meters from the ground. How many seconds later will the upward thrown ball hit the ground after the downward thrown ball? What is their final speed when they hit the ground? How many seconds will each ball be visible through the window? If the above building is 100 meters above a lake, which is 60 meters deep, and the balls sink at constant velocity upon entering the lake, how many total seconds will it take each ball to reach the bottom of the lake?

Explanation / Answer

For ball 1 height reached above the building v^2=u^2-2gh [v=0,u=20] =>h= 20.408 m time take to reach this height v=u-gt =>t= 2.04 sec It then falls from this height so total distance travelled downwards= 100+20.408=120.408m time taken by ball 1 to cover this s=u*t+0.5*g*t^2 [u=0] t=4.95 sec So total time taken by ball 1 = 2.04+4.95= 6.99sec Time taken by ball 2 s=u*t+0.5*g*t^2 [s=100,u=15] 4.9t^2+15t-100=0 =>t=3.24 sec A) 6.99-3.24=3.75sec B)For ball 1 v=u+gt [u=0,t= 4.95sec] v= 48.51 m/sec For ball 2 v=u+gt [u=15,t= 3.24] =>v=46.752m/sec C)For ball 1 distance coverd till top of window = 120.408-60= 60.408m velocity at this point v^2=u^2+2*g*h [=0,h= 60.04] =>v=34.30m/sec Length of window= 60-2.5=57.5m time it will be visible s=ut+0.5gt^2[ s=57.5,u= 34.3] =>4.9t^2+ 34.3t-57.5=0 =>t=1.39 sec [for ball 1] For ball 1 distance coverd till top of window = 100-60= 40 m velocity at this point v^2=u^2+2gh [h=40,u=15] =>v=31.764m/sec Length of window= 60-2.5=57.5m time it will be visible s=ut+0.5gt^2[ s=57.5,u= 31.764] =>4.9t^2+31.764t-57.5=0 =>t= 1.474sec [for ball 2] D)Distance covered by ball 1 till lake surface from its top point=120.408+100= 220.408 Velocity of ball 1 at lake surface v^2=u^2+2gh [u=0,h= 220.408] =>v=65.72m/sec Time taken to cover this distance v=u+gt[u=0] =>t= 6.706 sec Since ball moves with constant velocity after this for 60 m time taken by it is =60/65.72=0.9128 sec total time taken by ball 1 =2.04+6.706=8.74sec Distance covered by ball 2 till lake surface from its top point=100+100= 200 Velocity of ball 2 at lake surface v^2=u^2+2gh[u=15,h= 200] =>v=64.38 m/sec time taken by ball2 to reach this velocity v=u+gt[u=15,v=64.38] =>t=5.038 sec Since ball moves with constant velocity after this for 60 m time taken by it is 60/64.38=0.93194sec Total time taken= 5.038+0.93194=5.97sec

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