Question

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

• A. $30^{\circ }$
• B. $60^{\circ }$
• C. $45^{\circ }$
• D. $0^{\circ }$

Answer 1

Answer: (B)

Maximum height, $H=\frac{u^2{\sin }^2\theta }{2g}$
or $gH=\frac{u^2{\sin }^2\theta }{2}\dots (i)$
Velocity at the highest point, $v_H=u\cos \theta$
Let $v_x,v_y$ be the horizontal and vertical velocity of projectile at height $\frac{H}{2}.$ Then, $v_x=u\cos \theta$ and
$v_y^2=u^2{\sin }^2\theta -2g\times \frac{H}{2}=u^2{\sin }^2\theta -gH$
$=u^2{\sin }^2\theta -\frac{u^2{\sin }^2\theta }{2}=\frac{u^2{\sin }^2\theta }{2}$ (Using(i))
$\therefore$ Net velocity at height $\frac{H}{2}={\left(v_x^2+v_y^2\right)}^{1/2}$
As per question
$\sqrt{\frac{2}{5}}{\left(v_x^2+v_y^2\right)}^{1/2}=v_H$ or
$\frac{2}{5}\left(v_x^2+v_y^2\right)=v_H^2$
or $\frac{2}{5}\left[u^2{\cos }^2\theta +\frac{u^2}{2}{\sin }^2\theta \right]=u^2{\cos }^2\theta$
or ${\sin }^2\theta =3{\cos }^2\theta$ or $\sin \theta =\sqrt{3}\cos \theta$
or $\tan \theta =\sqrt{3}=\tan 60^{\circ }$ or $\theta =60^{\circ }$