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
