A body is projected vertically upward. Show the following.
(i) The distance covered by it in the first $n$ seconds of ascent is equal to the distance covered by it in the last $n$ seconds of its descent.
(ii) The distance covered by it in the last $n$ seconds of ascent is equal to the distance covered by it in the first $n$ seconds of its descent.

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Solution

Find the velocity $v_{1}$ of a body projected upwards after $t = (t_{1} - n)$ seconds
$t_{1} = t i m e o f a s c e n t = \frac{u_{1}}{g}$ using equation
$v_{1} = u_{1} - g t$

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Find the velocity v1 of a body projected upwards after t=(t1−n) secondst1=time of ascent=u1g using equationv1=u1−gt

Find the velocity $v_{1}$ of a body projected upwards after $t = (t_{1} - n)$ seconds
$t_{1} = t i m e o f a s c e n t = \frac{u_{1}}{g}$ using equation
$v_{1} = u_{1} - g t$