# The velocity of a projectile when it is at its highest point is√2/5 of its velocity when it is at half its greatest height. what is the angle of projection?

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by adityabhs

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by adityabhs

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H2 = 1/2 H1

velocity at highest point Vy = 0 (vertical component)

0 = Vo sinα - gt

t = Vo sinα / g

V1 = Vo cosα (horizontal velocity remain constant)

H1 = Vo sinα t - gt²/2

H1 = (Vo² - V1²) / 2g --------(1) (put value of t and solve)

velocity when it is at half of the greatest height

V2 = √[Vy² + Vx²]

V2 = √[(Vo sinα - gt)² + (Vo cosα)²]

Vo sinα t - gt² / 2 = (Vo² - V2²)/2g (just solve)

H2 = Vo sinα t - gt²/2 = (Vo² - V2²)/2g ----------(2)

H2 = 1/2 H1

substitute value of H1 & H2 and V1 = √(2/5) V2 then solve

V2² = 5/8 Vo²

put value of V2 and solve for V1

V1 = 1/2 Vo = Vo cosα

cosα = 1/2

α = 60

hence angle of projection = 60