Asoftball player hits a ball at
29.5 m/s at a 33.0° angle. an
outfielder catches the ball at the same height. how far was the
outfielder from the hitter?

81.1 m

Explanation:

Given:

x₀ = 0 m

y₀ = y = 0 m

v₀ₓ = 29.5 cos 33.0° m/s

v₀ᵧ = 29.5 sin 33.0° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

Find:

x

First, find the time it takes to land.

y = y₀ + v₀ᵧ t + ½ aᵧt²

0 = 0 + (29.5 sin 33.0) t - 4.9 t²

t = (29.5 sin 33.0) / 4.9

t = 3.28 s

Now in the x direction:

x = x₀ + v₀ₓ t + ½ aₓt²

x = 0 + (29.5 cos 33.0) (3.28) + 0

x = 81.1 m

The outfielder was 81.1 meters away.

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.