The angle of projection for a projectile to have same horizontal range and maximum height is :
- \(\tan^{-1}(2)\)
- \(\tan^{-1}(4)\)
- \(\tan^{-1}\left(\frac{1}{4}\right)\)
- \(\tan^{-1}\left(\frac{1}{2}\right)\)
The Correct Option is B
Approach Solution - 1
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:
\(\tan^{-1}(4)\)
Approach Solution -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:
\[\theta = \tan^{-1}(4).\]
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