how far will the ball go before being caught one meter above the ground?

32 meters

Explanation:

Given:

x₀ = 0 m

y₀ = 1 m

y = 1 m

v₀ₓ = 18 cos 52° m/s

v₀ᵧ = 18 sin 52° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

Find: x

First, find the time it takes for the ball to be caught.

y = y₀ + v₀ᵧ t + ½ aᵧt²

1 = 1 + 18 sin 52° t + ½ (-9.8) t²

0 = 14.2 t − 4.9 t²

0 = t (14.2 − 4.9 t)

t = 0 or 2.89

It takes 2.89 seconds to be caught.  The horizontal distance traveled in that time is:

x = x₀ + v₀ₓ t + ½ aₓt²

x = 0 + (18 cos 52°) (2.89) + ½ (0) (2.89)²

x = 32.1

Rounded to two significant figures, the ball travels 32 meters.

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