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°.