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miscellaneous
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A crewman on the starship Enterprise is on shore leave on a distant planet. He drops a rock from the top of a cliff and observes that it takes 3.00 s to reach the bottom. He now throws another rock vertically upwards so that it reaches a height of 2.00 m before dropping down the cliff face. The second rock takes 4.12 s to reach the bottom of the cliff. The planet has a very thin atmosphere that offers negligible air resistance. (a) How high is the cliff? (b) What is the value of g, the acceleration due to gravity, on the planet?
Here's how I did this:
2 + h = whole height which took 4.12 seconds
2+h = .5a(4.12)^2
h= 8.4872a - 2
h= .5a(3)^2
h= 4.5a
8.4872a-2 = 4.5a
a= .5 m/s^2
h=4.5a
h= 4.5(.5)
h= 2.257 m
Can anyone PLEASE verify these answers? Thanks a lottt
Here's how I did this:
2 + h = whole height which took 4.12 seconds
2+h = .5a(4.12)^2
h= 8.4872a - 2
h= .5a(3)^2
h= 4.5a
8.4872a-2 = 4.5a
a= .5 m/s^2
h=4.5a
h= 4.5(.5)
h= 2.257 m
Can anyone PLEASE verify these answers? Thanks a lottt