A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the…

Question

A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is ... (Assume, g=10 m s^-2 )

Accepted Answer

Correct answer: A. (a) : Time taken by the ball to reach highest point is t=1 s As the person throws the second ball, when the velocity of the first ball becomes zero, i.e., v=0 or when the first ball reach the highest point. Using, v=u+ at Here, v=0 ,a=-g ,t=1 s ∴ 0=u- (10 ) (1 ) u=10 m/s Using v^2-u^2=2 ah, we get (0 )^2- (10 )^2=2 (-10 ) (h )h= (10 )^220=5 m

$A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is ... (Assume, g = 10 m s^{- 2})$

Solution

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