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(A) The motion of a ball thrown vertically upwards is symmetric. The distance covered by the ball during the last $t$ seconds of its upward motion is

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

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(A) The motion of a ball thrown vertically upwards is symmetric. The distance covered by the ball during the last $t$ seconds of its upward motion is exactly equal to the distance covered by it in the first $t$ seconds of its downward motion starting from the highest point.At the highest point,the initial velocity for the downward motion is $u_{down} = 0$.Using the equation of motion $h = ut + \frac{1}{2}at^2$ for the downward journey:Here,$u = 0$,$a = g$,and time $= t$.Substituting these values,we get:$h = (0)t + \frac{1}{2}gt^2$$h = \frac{1}{2}gt^2$Therefore,the distance covered during the last $t$ seconds of its ascent is $\frac{1}{2}gt^2$.

If a ball is thrown vertically upwards with speed $u$,the distance covered during the last $t$ seconds of its ascent is

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