Question

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?

Answer 1

V1 = √(2/5) V2
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