Answer to Question #81291 in Mechanics | Relativity for Ravi yadav

Question

Two balls are intially seprated by 'd' if ball 2 is throwing down with 'u' after time t>v/g where v is the velocity of ball 1 which throwing upward find minimum value of u for which collision can occurs between balls

Answer 1

The first ball is thrown upward with speed v. After time t>v/g the second ball will be thrown down with speed u. Find minimum value of u for which collision may occur.
The second ball will be thrown down after the second ball has started falling down. If the second ball is thrown down with u=0, it will never reach the first ball since it will be falling already increasing its speed. At moment t=v/g the first ball reaches height
h=(gt^2)/2=v^2/2g,
stops going up and starts falling. In time t_1 after the ball have started falling its coordinates measured from the point of the second ball are:
y_1=h+(gt_1^2)/2=v^2/2g+(gt_1^2)/2.
Assume that there is a time delay τ in which the second ball is thrown. Its coordinates:
y_2=u(t_1-τ)+(g〖(t_1-τ)〗^2)/2.
The second ball will collide with the first when
y_1+y_2=d.
Thus
d-v^2/2g-(gt_1^2)/2=u(t_1-τ)+(g(t_1-τ)^2)/2,
u=1/(t_1-τ) (d-v^2/2g-g/2(2t_1^2+τ^2-2τt_1)).
Answer
u=1/(t_1-τ) (d-v^2/2g-g/2(2t_1^2+τ^2-2τt_1))