Question

A ball is thrown horizontally from the top of a 65 m building and lands 135 m from the base of the building. Ignore air resistance, and use a coordinate system whose origin is at the top of the building, with positive y upwards and positive x in the direction of the throw.(b) What must have been the initial horizontal component of the velocity in m/s?

Answer 1

37 m/s

Step-by-step explanation:

Given:

sₓ = 135 m

sᵧ = -65 m

uᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

Find: uₓ

First, find the time it takes to land.

sᵧ = uᵧ t + ½ aᵧ t²

-65 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 3.64 s

Next, find the horizontal velocity.

sₓ = uₓ t + ½ aₓ t²

135 m = uₓ (3.64 s) + ½ (0 m/s²) (3.64 s)²

uₓ = 37.1 m/s

Rounded to two significant figures, the horizontal velocity is 37 m/s.