Question

the speed of a projectile when it is at its greatest height is root(2/5) times its speed at half the max height . what is the angle of projection ?

Answer 1

Answer:

tan∧-1 (2) = 63.44°

Explanation:

the speed of a projectile when it is at its greatest height= ucos∝

1/2max height = u²sin²∝/4g = u²/4g = u²/40

u²/40 = utsin∝ - 5t²

5t² - utsin∝ + u²/40 = 0

t = usin∝ ± √ u²sin²∝ - u²/2 ÷ 10 = u/10{ sin∝±√sin²∝-0.5}

vy = u sin∝ -10t = usin∝ - u{ sin∝±√sin²∝-0.5} = ±u√sin²∝-0.5

v² = vx² + vy² = u²( sin²∝-0.5) + u²cos²∝ = u²( 1 - 0.5) = u²/2

v = u/√2

the speed of a projectile when it is at half max. height is = v = u/√2

now

ucos∝ = (√2/√5) u/√2 = u / √5

cos∝ = 1/√5

tan∝ = 2

∝ = tan∧-1 (2) = 63.44°