Hard

Question

The speed of projectile at its maximum height is half of its initai speed. The angle of projection is

Options

A.60o
B.15o
C.30o
D.45o

Solution


        
Let v be velocity of projectile at maximum height H
v = ucosθ
According to given problem, v =
∴ = cos θ         ⇒ cos θ =
⇒    θ = 60o 

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