Question

The angle of projection for a projectile to have same horizontal range and maximum height is :

• A. $\tan^{-1}(2)$
• B. $\tan^{-1}(4)$
• C. $\tan^{-1}\left(\frac{1}{4}\right)$
• D. $\tan^{-1}\left(\frac{1}{2}\right)$

Answer 1

Answer: (B)

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

Answer 2

The horizontal range is:
$\frac{u^2 \sin 2\theta}{g},$
and the maximum height is:
$\frac{u^2 \sin^2 \theta}{2g}.$
Equating:
$\frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{2g}.$
Simplifying:
$2 \sin \theta \cos \theta = \frac{\sin^2 \theta}{2}.$
Substitute $\sin 2\theta = 2 \sin \theta \cos \theta$:
$4 \sin \theta \cos \theta = \sin^2 \theta \implies 4 = \tan \theta.$
Thus:
$\theta = \tan^{-1}(4).$