Explanation:
Given:
x₀ = 0 m
y₀ = 0 m
v₀ = 20 m/s
θ = 35°
aᵧ = -9.8 m/s²
1) Find t when y = 0.
y = y₀ + v₀ᵧ t + ½ aᵧ t²
0 = 0 + (20 sin 35°) t + ½ (-9.8) t²
0 = t (20 sin 35° - 4.9 t)
t = 0, t = 2.34
The ball stays in the air 2.34 seconds.
2) Find y when vᵧ = 0.
vᵧ² = v₀ᵧ² + 2aᵧ (y - y₀)
0² = (20 sin 35)² + 2(-9.8) (y - 0)
y = 6.71 m
The ball reaches a maximum height of 6.71 meters.
3) Find x when y = 0.
x = x₀ + v₀ₓ t + ½ aₓ t²
x = 0 + (20 cos 35°) (2.34) + ½ (0) (2.3)²
x = 38.4 m
The ball lands 38.4 meters from Tom.
4) Find v when y = 0.
vₓ = aₓ t + v₀ₓ
vₓ = (0) (2.34) + 20 cos 35°
vₓ = 16.4 m/s
vᵧ = aᵧ t + v₀ᵧ
vᵧ = (-9.8) (2.34) + 20 sin 35°
vᵧ = -11.5 m/s
v = √(vₓ² + vᵧ²)
v = √((16.4)² + (-11.5)²)
v = 20 m/s
The ball has a speed of 20 m/s just before it lands.
