Question

A ball is thrown vertically upwards with a velocity $u$ from the ground. The ball attains a maximum height $H_{max.}$. Then find out the displacement at which ball has half of the maximum speed.

• A. $\frac{H_{max}}{2}$
• B. $\frac{3}{2}{H}_{max}$
• C. $$\frac{2}{3}{H}_{max}$$
• D. $$\frac{3}{4}{H}_{max}$$

Answer 1

The correct option is D $$\frac{3}{4}{H}_{max}$$

Let the speed of the ball is half i.e. $\frac{u}{2}$ at point $\text{B}$ at height $h$ as shown in figure.

Applying 3rd equation of motion, between the points $\text{A}$ and $\text{B}$,

${\left(\frac{u}{2}\right)}^2=u^2-2gh$

$\Rightarrow 2gh=\frac{3u^2}{4}$

$\Rightarrow h=\frac{3u^2}{8g}....\left(1\right)$

As the speed of the ball is zero at the maximum height $H_{max}$, so on applying equation of motion between points $\text{A}$ and $\text{C}$, we get

$0^2=u^2-2g{H}_{max}$

$\Rightarrow \frac{u^2}{2g}=H_{max}....\left(2\right)$

So, from eqs. (1) & (2),

$h=\frac{3}{4}\times \frac{u^2}{2g}=\frac{3}{4}{H}_{max}$

Hence, option $\left(d\right)$ is correct.