Question

(b) A ball is thrown upward from the top of a 30 m tower with initial velocity 90 m/s at an angle

O = 20°
Find the time to reach the ground
(Smarks)
Find the magnitude and direction of the velocity at the moment of impact (5 marks)?​

Answer 1

7.14 s

93.2 m/s, 24.9° below the horizontal

Step-by-step explanation:

Given in the y direction:

Δy = -30 m

v₀ = 90 m/s sin 20° ≈ 30.8 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-30 m = (30.8 m/s) t + ½ (-9.8 m/s²) t²

4.9t² − 30.8t − 30 = 0

t = [ 30.8 ± √((-30.8)² − 4(4.9)(-30)) ] / 2(4.9)

t = 7.14 s

Find: vᵧ

v² = v₀² + 2aΔy

vᵧ² = (30.8 m/s)² + 2 (-9.8 m/s²) (-30 m)

vᵧ = -39.2 m/s

The magnitude of the velocity is:

v² = vₓ² + vᵧ²

v² = (90 m/s cos 20°)² + (-39.2 m/s)²

v = 93.2 m/s

The direction of the velocity is:

tan θ = vᵧ / vₓ

tan θ = (-39.2 m/s) / (90 m/s cos 20°)

θ = -24.9°