2 balls are thrown up. One ball is thrown 2 seconds after the second ball. If the initial velocity is 40m/s. Find the height at which they will meet

Answers (1)
S safeer

Let the balls meet at time t after the first ball is thrown. Therefore the time for second ball is t-2

height is the same from the ground

\\40t-0.5\times10t^2=40(t-2)-0.5\times10(t-2)^2\\implies\ t=5sec

Therefore time taken by second ball t-2 = 5-2=3 sec

The first ball reaches the maximum height at t= 40/10 = 4sec

At time t = 4sec, the ball has reached the maximum height and while returning it meet with ball 2, which has travelled for 3 sec

Height travelled by the second ball in 3 sec is the meeting point from the ground and is given by

\\h= ut-0.5gt^2\\\\h=40\times3-0.5\times10\times3^2=75m

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