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A softball 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?

Respuesta :

MathPhys

Answer:

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.

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