How High is the Cliff and What is the Gravity on This Alien Planet?

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Homework Help Overview

The problem involves a scenario where a crewman drops and throws rocks from a cliff on an alien planet, aiming to determine the height of the cliff and the acceleration due to gravity on that planet. The context includes time measurements for the rocks' motions and the assumption of negligible air resistance.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the interpretation of the time measurements for the rocks, particularly whether the total time includes the ascent and descent of the thrown rock. There are questions about the necessary information to solve for the cliff height and gravity, as well as the implications of the given data.

Discussion Status

The discussion is ongoing, with some participants offering guidance on how to approach the problem by suggesting the formulation of relevant equations. There is a mix of interpretations regarding the information provided and its sufficiency for solving the problem.

Contextual Notes

Participants note the lack of information regarding the initial velocity of the thrown rock, which raises questions about the assumptions made in the problem setup. There is also a mention of the need for clarity on the time measurements related to the thrown rock.

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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
 
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miscellaneous said:
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.
I assume that the time of 4.12 s is the total time from when the rock is throw to when it hits the ground, not merely the time it took to fall from its maximum height to the ground.
 
I think that if that was the case, the problem would be virtually impossible without more information because we don't know at what acceleration the rock was thrown vertically and therefore couldn't tell the velocity-->we couldn't tell that time..Do you agree? If I'm wrong, please let me know.
 
I think you have all the info needed to solve for the height of the cliff and the acceleration.
 
Can you please guide me in the right direction? I don't know what I should do..
 
Start by writing every equation that fits the data:
(1) rock thrown rises to a height of 2 m
(2) rock thrown hits the ground in 4.12 s
(3) rock dropped hits the ground in 3 s
 

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