Question

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

• A. $\theta = \tan^{ - 1} \bigg( \frac{1}{ 4} \bigg)$
• B. $\theta = \tan^{ - 1} ( 4 )$
• C. $\theta = \tan^{ - 1} (2)$
• D. $\theta = 45^\circ$

Answer 1

Answer: (B)

$\theta = \tan^{ - 1} ( 4 )$

Answer 2

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 = $\frac{ u^2\, \sin^2\,\theta }{2g}$ According to question R = H $\therefore \frac{ u^2 \sin\ 2 \ \theta }{g} = \frac{ u^2 \, \sin^2 \, \theta }{2g}$ $\frac{2 u^2 \, \sin \theta \,cos\, \theta }{ g} = \frac{ u^2 \, \sin^2\, \theta }{ 2g}$ $\tan \theta = 4$ or $\theta = \tan^{ - 1} (4)$