Question

The muzzle velocity of a projectile is 1500 fps and the distance of the target is 10 miles. Find the angle of elevation of the gun.

Answer 1

Given info : The muzzle velocity of a projectile is 1500 foot/sec and the distance of the target is 10 miles.

To find : The angle of elevation of the gun is...

solution : velocity of projectile, u = 1500 f/s = 457.2 m/s

distance of the target, R = 10 miles = 16 km = 16000 m

now using formula,

Horizontal range, R = u²sin2θ/g

⇒16000 = (457.2)² × sin2θ/10

⇒160000/(457.2)² = sin2θ

⇒sin2θ = 0.7654

⇒2θ ≈ 50°

⇒θ = 25°

Therefore the angle of elevation of the gun is 25°.