The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{tan}^{- 1} (2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{tan}^{- 1} (4)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{π}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C ${tan}^{- 1} (4)$
We know,
$H = \frac{u^{2} {sin}^{2} θ}{2 g} R = \frac{u^{2} sin 2 θ}{g}$

As,
$H = R \frac{u^{2} {sin}^{2} θ}{2 g} = \frac{u^{2} sin 2 θ}{g} tan θ = 4$
$\Rightarrow θ = {tan}^{- 1} (4)$

Suggest Corrections