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 $t^{{}^{\prime }}$.
Then $t_1+2t^{{}^{\prime }}=t_2$
$t^{{}^{\prime }}=\left(\frac{t_2-t_1}{2}\right)\dots (i)$
Total time taken to reach point $C$
$T=t_1+t^{{}^{\prime }}$
$=t_1+\frac{t_2-t_1}{2}$
$=\frac{2t_1+t_2-t_1}{2}=\left(\frac{t_1+t_2}{2}\right)$
Maximum height attained
$H_{\max }=\frac{1}{2}g(T{)}^2$
$=\frac{1}{2}g{\left(\frac{t_1+t_2}{2}\right)}^2$
$=\frac{1}{2}g\cdot \frac{{\left(t_1+t_2\right)}^2}{4}$
$\Rightarrow H_{\max }=\frac{1}{8}g\cdot {\left(t_1+t_2\right)}^{2}m$