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

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 + 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

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

measuring things 1 litre of water