(800) Problem 6: Gravitational acceleration on the moon is one sixth of that on

Question

(800) Problem 6: Gravitational acceleration on the moon is one sixth of that on Earth. A ball released from rest above the surface falls from height d in a time of 0.75 seconds ©theexpertta.com 25% Part (a) Write an expression for the final velocity vrof the ball when it impacts the surface of the moon assuming it is dropped from rest. This expression should be in terms of only, g (the gravitational acceleration on Earth), and time, t. Grade Summary Deductions Potential 30% 70% 789 HOME Submissions Attempts remaining: 2 (30% per attempt) detailed view END 30% BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 30% deduction per hint. Hints remaining: 4 Feedback: 30% deduction per feedback. 25% Part (b) Calculate the final velocity, vr numerically in ms 25% Part (c) Calculate the height. Cl (in meters), from which the ball was dropped. 25% Part (d) From how high up would this ball need to be dropped on the earth, de (in meters), if it took the same time to reach the ground as it did on the moon?

Explanation / Answer

Given that,

amoon = g / 6

t = 0.75 s

(a)

Using kinematic equation,

v = u + at

v = 0 + at

vf = -gt / 6

(b)

vf = -9.8*0.75 / 6

vf = -1.225 m/s

(c)

FRom equation of motion,

df = di + ut + (1/2)at^2

0 = d - (1/2)*9.8 / 6 * (0.75)^2

d = 0.4593 m

(e)

at earth, a = g

d = (1/2)*9.8*(0.75)^2

d = 2.75 m

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