A skier goes down a slope and detaches from the ground moving in the horizontal direction with a speed of 25m / s. The slope has an inclination of 35 ° a) At what point does the skier make contact again with the ground?

107 m down the incline

Explanation:

Given:

v₀ₓ = 25 m/s

v₀ᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

-Δy/Δx = tan 35°

Find: d

First, find Δy and Δx in terms of t.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (0 m/s) t + ½ (-10 m/s²) t²

Δy = -5t²

Δx = v₀ₓ t + ½ aₓ t²

Δx = (25 m/s) t + ½ (0 m/s²) t²

Δx = 25t

Substitute:

-(-5t²) / (25t) = tan 35°

t/5 = tan 35°

t = 5 tan 35°

t ≈ 3.50 s

Now find Δy and Δx.

Δy ≈ -61.3 m

Δx ≈ 87.5 m

Therefore, the distance down the incline is:

d = √(x² + y²)

d ≈ 107 m

im not to sure what it is

f=ma newton's second law 7x8=56n