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Question:

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

60o

45o

30o

15o

Solution:

Let v be velocity of a projectile at maximum height H
v = ucosθ
According to given problem, v = u/2
∴ u/2 = ucosθ ⇒ cosθ = 1/2
θ = 60o