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Question

Question: The angle of projection at which the horizontal range and maximum height of projectile are equal is...

The angle of projection at which the horizontal range and maximum height of projectile are equal is

A

45o45^{o}

B

θ=tan1(0.25)\theta = \tan^{- 1}(0.25)

C

θ=tan14\theta = \tan^{- 1}4or (θ=76)(\theta = 76{^\circ})

D

60o60^{o}

Answer

θ=tan14\theta = \tan^{- 1}4or (θ=76)(\theta = 76{^\circ})

Explanation

Solution

R=4Hcotθ.R = 4H\cot\theta.

When R=HR = H then cotθ=1/4θ=tan1(4)\cot\theta = 1 ⥂ / ⥂ 4 \Rightarrow \theta = \tan^{- 1}(4)