Answer all the parts and show work. Will give brainliest. 35 points.

Explanation:

Given:

v₀ₓ = 65 m/s cos 50° = 41.78 m/s

aₓ = 0 m/s²

v₀ᵧ = 65 m/s sin 50° = 49.79 m/s

aᵧ = -9.8 m/s²

(a) Find t when vᵧ = 0 m/s.

vᵧ = aᵧ t + v₀ᵧ

0 m/s = (-9.8 m/s²) t + 49.79 m/s

t = 5.08 s

(b) Assuming the ball lands at the same height it was launched at, the total time in the air is double the time it takes to reach the maximum height.

t = 2 (5.08 s)

t = 10.2 s

(c) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2aᵧΔy

(0 m/s)² = (49.79 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 126 m

(d) Find Δx when t = 10.2 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (41.78 m/s) (10.2 s) + ½ (0 m/s²) (10.2 s)²

Δx = 425 m

change tension