Question

ERCISES A ball shot at an angle of 60° to the ground strikes a building 23m away at a point 16m high. Find the magnitude and direction of the final velocity of the ball as it strikes the wall. V 11 13m/s at 18 87° below the horizontal,​

Answer 1

Step-by-step explanation:

Given:

x₀ = 0 m

x = 23 m

y₀ = 0 m

y = 16 m

v₀ᵧ = v₀ₓ tan 60° = v₀ₓ √3

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

Find: vₓ, vᵧ

First, in the x direction:

Δx = v₀ₓ t + ½ aₓ t²

23 = v₀ₓ t + ½ (0) t²

t = 23 / v₀ₓ

Next, in the y direction:

Δy = v₀ᵧ t + ½ aᵧ t²

16 = v₀ᵧ t + ½ (-9.8) t²

16 = v₀ᵧ t − 4.9 t²

Substitute:

16 = (v₀ₓ √3) (23 / v₀ₓ) − 4.9 (23 / v₀ₓ)²

16 = 23√3 − 2592.1 / v₀ₓ²

2592.1 / v₀ₓ² = 23√3 − 16

v₀ₓ² = 2592.1 / (23√3 − 16)

v₀ₓ = 10.43 m/s

Now find v₀ᵧ and t:

v₀ᵧ = v₀ₓ √3

v₀ᵧ = 18.06 m/s

t = 23 / v₀ₓ

t = 2.206 s

Use these to find the final velocities.

vₓ = aₓ t + v₀ₓ

vₓ = (0) (2.206) + 10.43

vₓ = 10.43 m/s

vᵧ = aᵧ t + v₀ᵧ

vᵧ = (-9.8) (2.206) + 18.06

vᵧ = -3.55 m/s

Use Pythagorean theorem to find the magnitude:

v² = vₓ² + vᵧ²

v² = (10.43)² + (-3.55)²

v = 11.02 m/s

Use trig to find the direction:

tan θ = vᵧ / vₓ

tan θ = -3.55 / 10.43

θ = -18.82°