From the top of 100m high building a ball is dropped and at the same time another ball is thrown vertically upwards from the ground with a velocity of 24m/s, At what height will they cross each other? (Take g=10 m/s^2)
Let the ball dropped down is a and the ball thrown up is b. also the stone thrown upwards cover a distance of x and the other covers a distance of (100 –x). For ball a u=0 g=10m/s2 d=(100−x) Using the equation s=ut+21at2 100−x=5t2.........(1) For ball b d=x g=−10m/s2 u=25m/s s=ut+21at2 x=25t−5t2............(2) Solving equation (1) and (2) 100=25t t=4seconds Put the value of t in equation (1) x=100−80 x=20m They will meet at distance of 80 m from the ground after t = 4 seconds
