Question

A body is projected vertically upwards at time $t=0$ and it is seen at a height $H$ at time $t_{1}$ and $t_{2}$ second 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} $

Answer 1

Answer: (B)

Let time taken by the body to fall from point $C$ to $B$ is $t^{'}$.
Then $t_{1}+2 t^{'}=t_{2}$
$t^{'}=\left(\frac{t_{2}-t_{1}}{2}\right) \,\,\,\,\ldots( i )$
Total time taken to reach point $C$
$T=t_{1}+t^{'}$
$=t_{1}+\frac{t_{2}-t_{1}}{2}$
$=\frac{2 t_{1}+t_{2}-t_{1}}{2}=\left(\frac{t_{1}+t_{2}}{2}\right)$
Maximum height attained
$H_{\max }=\frac{1}{2} g(T)^{2}$
$=\frac{1}{2}\, g\left(\frac{t_{1}+t_{2}}{2}\right)^{2}$
$=\frac{1}{2} \,g \cdot \frac{\left(t_{1}+t_{2}\right)^{2}}{4}$
$\Rightarrow \,\, H_{\max }=\frac{1}{8}\, g \cdot\left(t_{1}+t_{2}\right)^{2} m$