A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 31 m high. When it hits the ground at the base of the cliff, the rock has a speed of 30 m/s.

a. Assuming that air resistance can be ignored, what is the initial speed of the rock?
b. What is the greatest height of the rock as measured from the base of the cliff?

a)  v₀ = 39.83 v / s , b)   y = 111.94 m

Explanation:

a) Let's use the kinematic relations for this problem

                v² = v₀² - 2 g y

The speed at the base of the cliff is 30 m / s, the height y = 31 m

               v₀² = v² + 2 g y

                v₀ =√ (30² + 2 9.8 31)

                v₀ = 39.83 v / s

b) at the point of maximum height the speed is zero

                     0 = v₀² - 2 g (y –y0)

                     y = y₀ + v₀² / 2g

                     .y = 31 + 39.83²/2 9.8

                      y = 111.94 m

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.

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