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(B) Let the initial velocity of the ball be $v_0 = 20 \; m s^{-1}$ in the upward direction.At the maximum height,the final velocity of the ball is $v

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) Let the initial velocity of the ball be $v_0 = 20 \; m s^{-1}$ in the upward direction.At the maximum height,the final velocity of the ball is $v = 0 \; m s^{-1}$.The acceleration due to gravity is $a = -g = -10 \; m s^{-2}$.Let $h$ be the maximum height reached by the ball above the point of projection.Using the kinematic equation $v^2 = v_0^2 + 2ah$,we have:$0^2 = (20)^2 + 2(-10)h$$0 = 400 - 20h$$20h = 400$$h = 20 \; m$.Therefore,the ball will rise $20 \; m$ above the point of projection.

$A$ ball is thrown vertically upwards with a velocity of $20 \; m s^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25.0 \; m$ from the ground. How high will the ball rise above the point of projection (in $; m$)? Take $g = 10 \; m s^{-2}$.

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