Question

# The velocity at the maximum height of projectile is half of its initial velocity projection. The angle of projection is

A
30o
B
45o
C
60o
D
76o

Solution

## The correct option is D $${ 60 }^{ o }$$Lets consider, $$u=$$ initial velocity of the projectile$$\theta=$$ angle of projectionAt maximum height,$$velocity=ucos\theta$$$$\dfrac{u}{2}=ucos\theta$$$$cos\theta=\dfrac{1}{2}=0.5$$$$\theta=cos^{-1}(0.5)$$$$\theta=60^0$$The correct option is C.Physics

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