Question

A juggler tosses a ball up in the air with initial speed u. At the instant it reaches its maximum height H, he tosses up a second ball with the same initial speed. The two balls will collide at a height
a)H/4
b) H/2
C) 3H/4
D) ✓3H/4
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Answer 1

Let the Displacement be S1 and S2 for the 2 balls

According to the question firstly the Juggler tosses a ball with speed ( u ) and the ball reaches the maximum height ( H ) and now that ball will fall downwards .

So ,, for the first ball S1 = 1/2×gt²

For second ball which is also thrown upward with velocity ( u )

S2 = ut - 1/2×gt²

=> S1 + S2 = H

→ 1/2×gt² + ut - 1/2×gt² = H

➡ ut = H. ........... ( i )

=>> H = u²/ 2g ........( ii )

Then = t = u/2g

putting these values in S2 so that we could know that where the balls will collide .

S2 = u×u/2g - 1/2g×u² /4g

=> 3u² ÷ 4g

➡ 3H/4 .