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

carla slowed down in the beginning of her trip, stopped, and then sped up

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )