Question:

The angle of projection for a projectile to have same horizontal range and maximum height is :

Updated On: Nov 23, 2024
  • \(\tan^{-1}(2)\)
  • \(\tan^{-1}(4)\)
  • \(\tan^{-1}\left(\frac{1}{4}\right)\)
  • \(\tan^{-1}\left(\frac{1}{2}\right)\)
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The Correct Option is B

Solution and Explanation

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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