A ball is gently dropped from a height of 20 | vedclass

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Question

(C) Let the final velocity with which the ball strikes the ground be $v$ and the time taken be $t$.Initial velocity of the ball,$u = 0 \, m s^{-1}$.Di

Vedclass · Accepted Answer

$v=20 \, m s^{-1}; t=2 \, s$

Vedclass · Answer

(C) Let the final velocity with which the ball strikes the ground be $v$ and the time taken be $t$.Initial velocity of the ball,$u = 0 \, m s^{-1}$.Distance or height of fall,$s = 20 \, m$.Downward acceleration,$a = 10 \, m s^{-2}$.Using the third equation of motion: $v^2 - u^2 = 2as$.$v^2 = 2as + u^2$.$v^2 = 2 	imes 10 	imes 20 + 0^2 = 400$.Therefore,final velocity $v = \sqrt{400} = 20 \, m s^{-1}$.Using the first equation of motion: $v = u + at$.$20 = 0 + 10 	imes t$.Therefore,time $t = 20 / 10 = 2 \, s$.

$A$ ball is gently dropped from a height of $20 \, m$. If its velocity increases uniformly at the rate of $10 \, m s^{-2}$,with what velocity will it strike the ground? After what time will it strike the ground?

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