An arrow is shot at 27.0° above the horizontal. Its velocity is 54 m/s, and it hits the target.

Explanation:

(a) What is the maximum height the arrow will attain?

Given:

v₀ᵧ = 54 sin 27.0° m/s = 24.5 m/s

vᵧ = 0 m/s

aᵧ = -9.8 m/s²

Find: Δy

vᵧ² = v₀ᵧ² + 2aᵧΔy

(0 m/s)² = (24.5 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 30.7 m

(b) The target is at the height from which the arrow was shot. How far away is it?

Given:

Δy = 0 m

v₀ᵧ = 54 sin 27.0° m/s = 24.5 m/s

aᵧ = -9.8 m/s²

Find: t

Δy = v₀ᵧ t + ½ aᵧt²

0 m = (24.5 m/s) t + ½ (-9.8 m/s²) t²

t = 5.00 s

Given:

v₀ₓ = 54 cos 27.0° m/s = 48.1 m/s

aₓ = 0 m/s²

t = 5.00 s

Find: Δx

Δx = v₀ₓ t + ½ aₓt²

Δx = (48.1 m/s) (5.00 s) + ½ (0 m/s²) (5.00 s)²

Δx = 241 m

the answer is a)

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