Question

gThe 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

Answer 1

Answer:

(a) 8 m/s

(b) 5 s

Explanation:

(a)

Using newton's equation of motion,

v² = u²+2gs ..................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity at the surface of moon, s = height reached.

make u the subject of the equation,

u² = v²-2gs

u = √(v²-2gs)................ Equation 2

Note: As the stone is thrown up, v = 0 m/s, g is negative

Given: v = 0 m/s, s = 20 m, g = -1.6 m/s²

Substitute into equation 2

u = √(0-2×20×[-1.6])

u = √64

u = 8 m/s.

(b)

Using,

v = u+ gt

Where t = time of flight to reach the maximum height.

Make t the subject of the equation,

t = (v-u)/g................................... Equation 3

Given: v =  0 m/s, u = 8 m/s, g = - 1.6 m/s²

Substitute into equation 3

t = (0-8)/-1.6

t = -8/-1.6

t = 5 seconds.