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(A) The total time taken by the ball to reach the maximum height is $T = \frac{u}{g}$.The motion in the last $t$ seconds of the upward journey is equi

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$\frac{1}{2}gt^2$

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(A) The total time taken by the ball to reach the maximum height is $T = \frac{u}{g}$.The motion in the last $t$ seconds of the upward journey is equivalent to the motion of a body falling freely from rest for $t$ seconds.Let the ball reach the maximum height at time $T = \frac{u}{g}$. At this point,its velocity is $v = 0$.Considering the motion in reverse (starting from the maximum height at $t=0$ with initial velocity $u_0 = 0$ and acceleration $g$ downwards),the distance $h$ covered in time $t$ is given by the kinematic equation:$h = u_0 t + \frac{1}{2} g t^2$Since $u_0 = 0$,we get:$h = 0 \cdot t + \frac{1}{2} g t^2 = \frac{1}{2} g t^2$Thus,the distance covered in the last $t$ seconds is $\frac{1}{2} g t^2$.

$A$ ball is thrown vertically upwards with a velocity $u$. What is the distance covered by it in the last $t$ seconds of its upward motion?

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