Question

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

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

Answer 1

Answer: (B)

${60}^0$

Time take to reach the max height $=\frac{u\sin \theta }{g}$

maximum height $=\frac{u^2{\sin }^2\theta }{2g}$

Half of the maximum height $=\frac{u^2{\sin }^2\theta }{4g}$

$u_x=u\cos \theta$ (at half height)

$u_y=\sqrt{u^2{\sin }^2\theta -2g\left(\frac{u\sin \theta }{2g}\right)}$

$=\frac{u\sin \theta }{\sqrt{2}}$

$v=\sqrt{u^2{\cos }^2\theta +\frac{u^2{\sin }^2\theta }{2}}$

$u\cos \theta =\sqrt{\frac{2}{5}}\sqrt{u^2{\cos }^2\theta +\frac{u^2{\sin }^2\theta }{2}}$

$\Rightarrow u^2{\cos }^2\theta =\frac{2}{5}{u}^2(1-{\sin }^2\theta +\frac{{\sin }^2\theta }{2})$

$\Rightarrow 5(1-{\sin }^2\theta )=2{\sin }^2\theta$

$\Rightarrow 3=4{\sin }^2\theta$

$\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}$

$\theta =\frac{\pi }{3}$

So, the correct answer is ${60}^{\circ }$