Question

A baseball player throws a baseball with a velocity of 13 m/s north. It is caught by a second player seven seconds later. How far is the second player from the first player?
A. 91 meters north
B. 91 meters south
C. 91 meters east
D. 91 meters west

Answer 1

Answer:

A. 91 meters north

Explanation:

Take +y to be north.

Given:

v₀ = 13 m/s

a = 0 m/s²

t = 7 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (13 m/s) (7 s) + ½ (0 m/s²) (7 s)²

Δy = 91 m

The displacement is 91 m north.