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. $\frac{g \, t_2 - {t_1}^2}{8}$
• B. $\frac{g \, t_2 + {t_1}^2}{4}$
• C. $\frac{g \, t_2 + {t_1}^2}{8}$
• D. $\frac{g \, t_2 - {t_1}^2}{4}$

Answer 1

Answer: (C)

$\frac{g \, t_2 + {t_1}^2}{8}$

Answer 2

Let time taken by the body to fall from point $C$ to $B$ is $t$'.
Then $t_1 + 2t' = t_2$
$t' = \frac{t_2 - t_1}{2}$ ....(i)
Total time taken to reach point C
$T = t_1 + t'$
$= t_1 + \frac{t_2 - t_1}{2}$
$= \frac{2 t_1 + t_2 - t_1}{2}$
$= \frac{t_1 + t_2} {2}$
Maximum height attained
$H_{max } = \frac{1}{2} \, g \, T^2$
$= \frac{1}{2} g \frac{t_1 + t_2}{2}^2$
$= \frac{1}{2} g . \frac{t_2 + t_2}{4}^2$
$\Rightarrow \, \, H_{max} = \frac{1}{8} g . t_1 + {t_2 }^2 m$