The angle is calculated and found to be θ = tan-1 (sqrt(6)) ≈ 67.8°.
To find the launching angle θ
tan (θ) = v0y/v0x, where v0y and v0x are the y- and x-components of the initial velocity whose magnitude is
v0= sqrt(v0y²+v0x²).
use kinematic equation that which does not contain time,
v² = v0² - 2gy = v0y²+v0x² - 2gy.
At the maximum height, y=h, the velocity only has an x-component, v=vx, since vy=0. Since there is no acceleration in the x-direction (assuming no air resistance), v=v0x. Use this into the kinematic equation and get
v0x² = v0y²+v0x² - 2gh ⇒ h = v0y2/2g.
Use the condition we are given:
vy=h = (1/2) vy=h/2, or v²y=h = (1/4) v²y=h/2.
This gives us
v0x² = (1/4) (v0x² +v0y² - 2g(h/2)) = (1/4) (v0x² +v0y² - g(v0y2/2g)) = (1/4) (v0x² + (1/2) v0y²).
Bring all v0x² terms to one side and divide by v0y², take the reciprocal and get
(v0y/v0x)² = 6, so that
tan (θ) = sqrt(6),
θ = tan-1 (sqrt(6)) ≈ 67.8°.
Learn more about the Projectile with the help of the given link:
https://brainly.com/question/13388411
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