1. Tardigrade
  2. Question
  3. Physics
  4. The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

AIPMTAIPMT 2012Motion in a Plane

Solution:

Horizontal range, $R=\frac{u^{2} \sin 2 \theta}{g}$
where $u$ is the velocity of projection and $\theta$ is the angle of projection
Maximum height, $H=\frac{u^{2} \sin ^{2} \theta}{2 g}$
According to question $R=H$
$\therefore \frac{u^{2} \sin 2 \theta}{g}=\frac{u^{2} \sin ^{2} \theta}{2 g} $
$\frac{2 u^{2} \sin \theta \cos \theta}{g}=\frac{u^{2} \sin ^{2} \theta}{2 g} $
$\tan \theta=4$
or $ \theta=\tan ^{-1}(4)$