Q. A body is projected vertically upwards at time $t = 0$ and it is seen at a height $H$ at instants $t_{1}$ and $t_{2}$ seconds during its flight. The maximum height attained is $(g$ is acceleration due to gravity$)$
A) $\frac{g\left( t_{2} - t_{1} \right)^{2}}{8}$
B) $\frac{g\left( t_{1} + t_{2} \right)^{2}}{4}$
C) $\frac{g\left( t_{1} + t_{2} \right)^{2}}{8}$
D) $\frac{g\left( t_{2} - t_{1} \right)^{2}}{4}$
Explanation
(c) Let time taken by the body to fall from point $C$ to $B$ be $t''$ Then, $t_{1} + 2t^{''} = t_{2}$
$t^{''} = \left( \frac{t_{2} - t_{1}}{2} \right)$ …(i)
Total time taken, to reach point $C$
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$T = t_{1} + t''$
$= t_{1} + \frac{t_{2} - t_{1}}{2}$
$= \frac{2t_{1} + t_{2} - t_{1}}{2} = \left( \frac{t_{1} + t_{2}}{2} \right)$
Then maximum height attained
$H_{\max} = \frac{1}{2}g(T)^{2} = \frac{1}{2}g\left( \frac{T_{1} + t_{2}}{2} \right)^{2}$
$= \frac{1}{2}g.\frac{\left( t_{1} + t_{2} \right)^{2}}{4}$
$\Rightarrow H_{\max} = \frac{1}{8}g.\left( t_{1} + t_{2} \right)^{2}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.