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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top