Question

The angle of projection for which the horizontal range is equal to the maximum height of a projectile is :

• A. $\theta ={\tan }^{-1}(1)$
• B. $\theta ={\tan }^{-1}(4)$
• C. $\theta ={\tan }^{-1}(3)$
• D. $\theta ={\tan }^{-1}(2)$

Answer 1

Answer: (B)

Let a particle be projected with initial velocity u at an angle $\theta$ with the horizontal.
Then, range $(R)$ of projectile is
$R=\frac{u^2\sin 2\theta }{g}\dots$ (i)
Maximum height $H=\frac{u^2{\sin }^2\theta }{2g}\dots$ (ii)
Given, $R=H$,
$\therefore \frac{u^2\sin 2\theta }{g}=\frac{u^2{\sin }^2\theta }{2g}$
$\Rightarrow 2\sin \theta \cos \theta =\frac{{\sin }^2\theta }{2}$
$\Rightarrow 4\cos \theta =\sin \theta$
$\Rightarrow \tan \theta =4$
$\Rightarrow \theta ={\tan }^{-1}(4)$