Question

The horizontal range and maximum height for a projectile is same. The angle of projectile is

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

Answer 1

The correct option is B $\theta ={\tan }^{-1}\left(4\right)$

Let angle of projection be $\theta$
We know,
Maximum height = $\frac{u^2{\sin }^2\theta }{2g}$
Horizontal range = $\frac{u^2\sin 2\theta }{g}$
According to question

$\frac{u^2{\sin }^2\theta }{2g}=\frac{u^2\sin 2\theta }{g}$
$\frac{{\sin }^2\theta }{2}=2\sin \theta \cos \theta$
$\tan \theta =4$
$\theta ={\tan }^{-1}\left(4\right)$