Example ProblemS 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignored, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff

initial speed 38.31 m/s

greatest height of the rock 106.88 m

Explanation:

given data

mass = 0.26 kg

high = 32 m

speed = 29 m/s

solution

we use here kinematic relations for this problem that is

v² = u² - 2 g S   1

here put the value we get and we get initial speed u

u² = v² + 2 g S

u² = 29² + 2×9.8×32

solve it we get

u = 38.31 m/s

and

at the point of maximum height the speed is zero so

v = u² - 2 g (S - S(0))   2

S = S(0) + u²  ÷ 2 g

put here value we get

S = 32 + 38.31²  ÷ 2 (9.8)  

S = 106.88 m

your answer is, orbit around the nucleus