Question

A cannon launches a projectile at an angle of 22 degrees off the horizontal​. If the initial velocity is 830m/s, find how high the projectile is launched?

Please give clear steps

Answer 1

Answer:

4900 m

Explanation:

Given, in the y direction:

v₀ = 830 sin 22° m/s ≈ 311 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (311 m/s)² + 2 (-9.8 m/s²) Δy

Δy ≈ 4930 m

Rounded to two significant figures, the vertical displacement is 4900 m.