Question

A student fires a cannonball horizontally with a speed of 39m/s from a height of

Answer 1

Explanation:

The initial horizontal speed is 39 m/s, and the initial vertical speed is 0 m/s.

v₀ₓ = 39 m/s

v₀ᵧ = 0 m/s

In the y direction:

Δy = 73 m

v₀ᵧ = 0 m/s

aᵧ = 9.8 m/s²

Find: t

Δy = v₀ᵧ t + ½ aᵧt²

73 = 0 + ½ (9.8) t²

t = 3.9 s

In the x direction:

v₀ₓ = 39 m/s

aₓ = 0 m/s²

t = 3.9 s

Find: Δx

Δx = v₀ₓ t + ½ aₓt²

Δx = (39) (3.9) + 0

Δx = 150 m