The acceleration of gravity at the surface of Moon is 1.6 m/s2. A 5.0 kg stone thrown upward on Moon reaches a height of 20 m. (a) Find its initial velocity. (b) What is the time of flight to reach the max height

(a) 8 m/s

(b) 5 s

Explanation:

(a)

Using,

V² = U²+2gh ... Equation 1

Where V = final velocity, U = Initial velocity, g = acceleration due to gravity on the surface of the moon, h = height reached.

Given: V = 0 m/s (At it's maximum height), g = - 1.6 m/s² (as its moves against gravity), h = 20 m.

Substitute into equation 1

0 = U²+[2*20 * (-1.6) ]

-U² = - 64

U² = 64

U = √64

U = 8 m/s.

(b)

V = U + gt ... Equation 2

Where t = time to reach the maximum height.

Given: V = 0 m/s (At the maximum height), g = - 1.6 m/s² (Moving against gravity), U = 8 m/s.

Substitute into equation 2

0 = 8 + (-1.6t)

-8 = - 1.6t

-1.6t = - 8

t = - 8/-1.6

t = 5 s.