Question

A body is projected vertically upwards. The times corresponding to height $h$ while ascending and while descending are $t_1$ and $t_2$ respectively. Then the velocity of projection is ($g$ is acceleration due to gravity).

• A. $g\sqrt{t_1{t}_2}$
• B. $\frac{g{t}_1{t}_2}{t_1+t_2}$
• C. $\frac{\sqrt{g{t}_1{t}_2}}{2}$
• D. $\frac{g(t_1+t_2)}{2}$

Answer 1

Answer: (D)

Let $v$ be initial velocity of vertical projection and $t$ be the time taken by the body to reach a height $h$ from ground.
Here, $u=u,a=-g,s=h,t=t$
Using, $s=ut+\frac{1}{2}a{t}^2$, we have
$\begin{array}{l}h=ut+\frac{1}{2}(-g)t^2\text{ or }g{t}^2-2ut+2h=0\\ \therefore t=\frac{2u\pm \sqrt{4u^2-4g\times 2h}}{2g}=\frac{u\pm \sqrt{u^2-2gh}}{g}\end{array}$
It means t has two values, i.e.,
$\begin{array}{l}{t}_1=\frac{u+\sqrt{u^2-2gh}}{g}\\ t_2=\frac{u-\sqrt{u^2-2gh}}{g}\\ t_1+t_2=\frac{2u}{g}\text{ or }u=\frac{g\left(t_1+t_2\right)}{2}\end{array}$