Question

A skier rides horizontally off of a 200 meter high cliff. If he lands 25 meters away from the base of the cliff, how fast was he skiing as he went off the edge?

A. 0.91 m/s
B. 2.91 m/s
C. 3.91 m/s
D. 25.91 m/s

Answer 1

Answer:

the answers, the correct one is C,   v₀ₓ = vₓ = 3.91 m / s

Explanation:

This is a projectile launching exercise, in this case they indicate that when leaving the cliff it goes horizontally, therefore the initial vertical speed is zero, let's find the time to reach the base

        y = y₀ + $v_{oy}$ t - ½ g t²

at the base the height is zero (y = 0 m)

        0 = y₀ + 0 - ½ g t²

         t = √ (2y₀ / g)

         

we calculate

        t = √ (2 200 / 9.8)

        t = 6.389 s

with this time we calculate the horizontal speed

        v₀ₓ = x / t

        v₀ₓx = 25 / 6,389

        v₀ₓ = vₓ = 3.91 m / s

When checking the answers, the correct one is C