At 1/2 its maximum height, the speed of a projectile is 3/4 of its initial speed. what was its launch angle?

In the projectile motion, the direction is characterized by a shape of the arc. Its horizontal component travels in constant velocity while the vertical component travels in constant acceleration. The equation to be used is:

2ay = |v² - v₀²|
where
a is the acceleration due to gravity equal to 9.81 m/s²
y is the height
v is the final velocity
v₀ is the initial velocity

Substitute y=1/2*H and v = 3/4*v₀. The equation for maximum height is

H = v₀²sin²θ/2a

Thus,

(2)(9.81)(1/2)(H) = |(3/4v₀)² - v₀²|
(2)(9.81)(1/2)(v₀²sin²θ/2(9.81)) = |(3/4v₀)² - v₀²|
(1/2)v₀²sin²θ = 7/16 * v₀₂
(1/2)sin²θ = 7/16
sin θ = [2*(7/16)]² = 0.765625
θ = sin⁻¹(0.765625) = 49.96°

Therefore, the launch angle is 49.96°.

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