Question

If the horizontal range of a projectile is equal to the maximum height reached, then the corresponding angle of projection is

• A. ${\tan }^{-1}$$\left(1\right)$
• B. ${\tan }^{-1}\left(\sqrt{3}\right)$
• C. ${\tan }^{-1}\left(4\right)$
• D. ${\tan }^{-1}\left(12\right)$

Answer 1

The correct option is C

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

Step 1: Given,

A projectile's horizontal range is equal to its maximum height reached.

Step 2: Calculating the angle of projection,
As we know that projectile motion
Horizontal limit range $=\frac{u^2\sin 2\theta }{g}$

And maximum height$=\frac{u^2{\sin }^2\theta }{2g}$

In the above formula, u sin(θ) is the component of initial velocity along the y-axis, u is the initial velocity and g is the acceleration due to gravity
As asked that Horizontal Range = Maximum Height
$\Rightarrow \frac{u^2\sin 2\theta }{g}=\frac{u^2{\sin }^2\theta }{2g}$
$\Rightarrow 2\sin \theta \cos \theta =\frac{{\sin }^2\theta }{2}$
$\Rightarrow 4\cos \theta =\sin \theta$
$\Rightarrow \tan \theta =4$
$\Rightarrow \theta ={\tan }^{-1}4$

Thus, the corresponding angle of projection is ${\tan }^{-1}4$.

Hence, option C is the correct answer.