Q. A stone is projected at an angle of 30° with the vertical. If the horizontal component of its velocity
is 9.8 m/s. Calculate the i) maximum height and
horizontal range

1.63 m

11.3 m

Explanation:

Given:

v₀ₓ = 9.8 m/s

v₀ᵧ = 9.8 tan 30° = 5.67 m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

Find: Δy given vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2aᵧΔy

(0 m/s)² = (5.67 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 1.63 m

Find: Δx given Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧt²

0 m = (5.67 m/s) t + ½ (-9.8 m/s²) t²

t = 1.15 s

Δx = v₀ₓ t + ½ aₓt²

Δx = (9.8 m/s) (1.15 s) + ½ (0 m/s²) (1.15 s)²

Δx = 11.3 m