Question

A body is thrown vertically upward 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. (9/4)h
• B. (4/3)h
• C. (9/8)h
• D. none of these

Answer 1

Answer: (C)

$H = \frac{u^{2}}{2g} \,\, ...\left(i\right) $
$ \frac{t_{1}}{t_{2} } = \frac{1}{2} $
$\Rightarrow 2t_{1 } =t_{2} \,\, ...\left(ii\right) $
$ u =\frac{1}{2}g\left(t_{1}+t_{2}\right) \,\, ...\left(iii\right)$
$h = \frac{1}{2}gt_{1}t_{2} \,\,...\left(iv\right) $
From $\left(ii\right), \left(iii\right)$ and $\left(iv\right)$
$u^{2} = \frac{9gh}{4}$
Put $u^{2}$ in equation $\left(i\right) : H = \frac{9}{8}h$