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

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.

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