Question

The speed of a projectile when it is at its maximum height is $\sqrt{\frac{2}{5}}$ times its speed at half the maximum height. The angle of projection is

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${45}^{\circ }$
• D. ${\tan }^{-1}\frac{3}{4}$

Answer 1

The correct option is B ${60}^{\circ }$

Let $\theta$ be the angle of projection and $u$ its initial speed. Then maximum height will be
$H=\frac{u^2{\sin }^2\theta }{2g}$
$\Rightarrow gH=\frac{u^2{\sin }^2\theta }{2}$
Now, $v_H^2=u^2{\cos }^2\theta ..........\left(1\right)$
Using 3rd equation of motion when the projectile is at half the maximum height,
$v_{H/2}^2=u^2-2g\frac{H}{2}=u^2-gH$
$=u^2-\frac{u^2{\sin }^2\theta }{2}.......\left(2\right)$
Now it is given that
$v_H=\sqrt{\frac{2}{5}}{v}_{H/2}$
$or{v}_H^2=\frac{2}{5}{v}_{H/2}^2$
Substituting the values from Eqn. $\left(1\right)$ and $\left(2\right)$, we get
$u^2{\cos }^2\theta =\frac{2}{5}[u^2-\frac{u^2{\sin }^2\theta }{2}]\Rightarrow 5{\cos }^2\theta =2[1-\frac{{\sin }^2\theta }{2}]$
$\Rightarrow 5(1-{\sin }^2\theta )=2-{\sin }^2\theta$
$\Rightarrow {\sin }^2\theta =\frac{3}{4}$
$\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}\text{or}\theta ={60}^{\circ }$