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

The Correct Option is B

To solve this question, we need to determine the angle of projection such that a projectile achieves the same horizontal range and maximum height.

In projectile motion, the horizontal range \( R \) and maximum height \( H \) are given by the following equations:

• \(R = \frac{v^2 \sin(2\theta)}{g}\)
• \(H = \frac{v^2 \sin^2(\theta)}{2g}\)

where:

• \( v \) = initial velocity of the projectile
• \( \theta \) = angle of projection
• \( g \) = acceleration due to gravity

 

We want \(R = H\).

Setting the equations equal gives:

\(\frac{v^2 \sin(2\theta)}{g} = \frac{v^2 \sin^2(\theta)}{2g}\)

Cancel the common terms and rearrange:

\(2 \sin(2\theta) = \sin^2(\theta)\)

Using the trigonometric identity \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\), we have:

\(2 \times 2\sin(\theta)\cos(\theta) = \sin^2(\theta)\)

Simplify to:

\(4 \sin(\theta) \cos(\theta) = \sin^2(\theta)\)

Divide both sides by \(\sin(\theta)\) (assuming \(\theta \neq 0\)):

\(4 \cos(\theta) = \sin(\theta)\)

This gives:

\(\frac{\sin(\theta)}{\cos(\theta)} = 4\)

So, \(\tan(\theta) = 4\).

The angle \(\theta\) is thus \(\theta = \tan^{-1}(4)\).

Therefore, the correct angle of projection is \(\tan^{-1}(4)\), making the correct answer the option:

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

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:
\[\theta = \tan^{-1}(4).\]