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A juggler tosses a a ball up in the air with initial speed u. at instant it reaches the maximum height H. he toses up a second ball with same initial Speed. the two balls will collide at a height.1. H/42.H/23. 3H/44. (3H/4)^1/2no spam and plz. plz solve in detail.Detailed ans will be marked as brainliest

Answer»

Heya......!!!

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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 .

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Hope It helps You. ☺


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