Question

A ball is projected from the ground with a velocity $u$ such that its range is maximum.

• A. Its velocity at half the maximum height is $\frac{\sqrt{3}}{2}u$.
• B. Its velocity at the maximum height is $u$.
• C. Change in its velocity when it returns to the ground is $u$.
• D. All the above are true.

Answer 1

The correct option is A Its velocity at half the maximum height is $\frac{\sqrt{3}}{2}u$.

Range, $R=\frac{u^2\sin 2\theta }{g}$
R is max. for $\theta ={45}^o$

$H_{max}=\frac{u^2{\sin }^2\theta }{2g}=\frac{u^{2}si{n}^2{45}^0}{2g}=\frac{u^2}{4g}$

So, at $H=\frac{H_{max}}{2}=\frac{u^2}{8g}$

$v_y^2=u_y^2-2gH$
$v_y^2=(u\sin {45}^o{)}^2-2g\times \frac{u^2}{8g}=\frac{u^2}{2}-\frac{u^2}{4}$ $=\frac{u^2}{4}$
$v_x=ucos{45}^o$ $=\frac{u}{\sqrt{2}}$
$\Rightarrow v_y=\frac{u}{2}$

Hence, $v=\sqrt{v_x^2+v_y^2}=\frac{\sqrt{3}u}{2}$