Q. The horizontal range of an oblique projectile is equal to the distance through which a projectile has to fall freely from rest to acquire a velocity equal to the velocity of projection in magnitude. The angle of projection is
A) $75{^\circ}$
B) $60{^\circ}$
C) $45{^\circ}$
D) $30{^\circ}$
Explanation
(a) Using $v^{2} - u^{2} = 2as$, we get
$s = \frac{v^{2}}{2g}$
Now, $\frac{v^{2}\sin{2\theta}}{g} = \frac{v^{2}}{2g}$ or $sin\ 2\theta = \frac{1}{\mathbf{2}}$
or $\sin{2\theta} = \sin{30{^\circ}}$ or $\theta = 15{^\circ}$
The other possible angle of projection is $(90{^\circ} - 15{^\circ})$,
$ie,\ 75{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.