Question

a projectile is thrown with velocity v at an angle theta with horizontal. it's maximum height is 8 meters and horizontal range is 24 meters. calculate v in terms of acceleration due to gravity ​

Answer 1

Answer:

v = 2√(5g) m/s

Explanation:

let theta = ∝

H = 8 = v²sin²∝/2g

R = 24 = v²sin∝/g

now

v²sin²∝ = 16g

v²sin2∝ = 24g

dividing

tan∝ = 4/3 , sin∝ = 4/5

v² = 16g/0.8 = 20g

v = 2√(5g) m/s