Question

A body is thrown vertically upwards with velocity $u$. The ratio of times at which it is at particular height $h$ is $1:2$. What is the maximum height reached by the body above point of projection?

• A. $\frac{9}{4}h$
• B. $\frac{4}{3}h$
• C. $\frac{9}{8}h$
• D. None of these

Answer 1

Answer: (C)

$\frac{9}{8}h$

$h=ut-\frac{1}{2}g{t}^2$ ..........$\left(1\right)$ (upward direction -g)

$g{t}^2-2ut+2h=0$

$t_1+t_2=\frac{2u}{g}$ $[\frac{t_1}{t_2}=\frac{1}{2}]$

$t_1+2t_1=\frac{2u}{g}$

$3t_1=\frac{2u}{g}$

$t_1=\frac{2u}{3g}$

For maximum height $H=\frac{u^2}{2g}$

$h=u\times \frac{2}{3}.\frac{u}{g}-\frac{1}{2}.g.\frac{4}{9}\frac{u^2}{g^2}$

$h=\frac{2}{3}\frac{u^2}{g}-\frac{2u^2}{9g}$

$h=\frac{6u^2-2u^2}{9g}$

$h=\frac{4u^2}{9g}$

$u^2=\frac{9gh}{4}$

$H=\frac{u^2}{2g}$

$H=\frac{9gh}{4}\times \frac{1}{2g}$

$H=\frac{9}{8}h$.