Question

A body is projected vertically upward with a velocity of $10m{s}^{-1}$. It reaches maximum height $h$ at time $t$. In time $\frac{t}{2}$, the height covered is

• A. $\frac{h}{2}$
• B. $\frac{2}{5}h$
• C. $\frac{3}{4}h$
• D. $\frac{5}{8}h$

Answer 1

The correct option is C $\frac{3}{4}h$

Given,

$Velocity=10m/s,$

Height $=h$

Time $=t$

Since $v=u-gt$ where $u=0$

$\Rightarrow 10=10t\Rightarrow t=1s$

Since $u=0$

At $t=\frac{1}{2}s$ $u^2=2\times gH$

$\Rightarrow \frac{100}{2\times 10}=H\Rightarrow H=5m$

Now, $h^{\prime }=ut+\frac{1}{2}g{t}^2=10\times \frac{1}{2}-\frac{1}{2}\times 10\times \frac{1}{4}=\frac{15}{4}=\frac{3H}{4}$