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

  • Thread starter Thread starter miscellaneous
  • Start date Start date
  • Tags Tags
    Rock
AI Thread Summary
A crewman on the starship Enterprise drops a rock from a cliff, observing it takes 3.00 seconds to reach the bottom, while a second rock thrown upwards reaches a height of 2.00 meters and takes 4.12 seconds to fall. The calculations suggest that the height of the cliff is approximately 2.257 meters, with the acceleration due to gravity on the planet determined to be 0.5 m/s². There is a discussion about whether the 4.12 seconds includes the time taken to rise and fall, indicating a need for clarity on the motion of the second rock. The participants agree that all necessary information is available to solve the problem. The thread seeks verification and guidance on the calculations presented.
miscellaneous
Messages
10
Reaction score
0
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
 
Physics news on Phys.org
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
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top