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

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Solution

Q.
The angle of projection for a projectile to have same horizontal range and maximum height is [2024]

1 $\tan^{- 1} (2)$

2 $\tan^{- 1} (\frac{1}{2})$

3 $\tan^{- 1} (4)$

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

Ans.
(3)

   $H_{m a x} = R a n g e$

$\frac{u^{2} \sin 2 θ}{g} \frac{u^{2} \sin^{2} θ}{2 g}$

   $4 \sin θ \cos θ = \sin^{2} θ$

   $\tan θ = 4 \Rightarrow θ = \tan^{- 1} 4$

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