A rock is thrown off of a cliff

  • Thread starter Gustason
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  • #1
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Homework Statement


A stone is thrown vertically upward with a speed of 14.0 m/s from the edge of a cliff 95.0 m high
(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting?
(c) What total distance did it travel?

Homework Equations


Well I'm not quite sure what equations to use, thats the problem.


The Attempt at a Solution


On this problem I'm confused about what equation to use in order to find these answers. How do I factor in that upward velocity of 14.0 m/s? I know that my acceleration is -9.8 so do I just take 14t-9.8t? How do you know the speed just before hitting or what total distance it traveled?
 

Answers and Replies

  • #2
You have studied the kinematic equations, haven't you? Your solution will require these.
 
  • #3
LowlyPion
Homework Helper
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Homework Statement


A stone is thrown vertically upward with a speed of 14.0 m/s from the edge of a cliff 95.0 m high
(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting?
(c) What total distance did it travel?

Homework Equations


Well I'm not quite sure what equations to use, thats the problem.

The Attempt at a Solution


On this problem I'm confused about what equation to use in order to find these answers. How do I factor in that upward velocity of 14.0 m/s? I know that my acceleration is -9.8 so do I just take 14t-9.8t? How do you know the speed just before hitting or what total distance it traveled?

The first question relates to time. You will need to use the knowns that you have like initial velocity, g, and distance and relate those to time and solve.

In the second part you can use the equation that relates initial velocity to final velocity, because you do know the other needed variables.

The last part merely want you to calculate how high it went, double it and add it to 95. (Hint: At the top of it's flight its velocity is 0, so knowing initial velocity ...)

Some useful equations:
https://www.physicsforums.com/showpost.php?p=905663&postcount=2
 
  • #4
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I answer your first question you would use the motion equation position vs time:
14m*t+1/2(9.8)t
14/4.9=2.857 sec
you will have to use the secs found to find the other part of the equation I found the point from it going up to being parellel with the cliff, now you have to take it further and find how much time it takes to hit the bottum.
 
  • #5
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Hmm I tried using your method kmiller and I got the answer incorrect. I took the 2.857 seconds and added them to the 19.38 (95/4.9) seconds that the rock would fall during the 95m portion and I came up with 22.245 seconds.
 
  • #6
LowlyPion
Homework Helper
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I answer your first question you would use the motion equation position vs time:
14m*t+1/2(9.8)t
14/4.9=2.857 sec
you will have to use the secs found to find the other part of the equation I found the point from it going up to being parellel with the cliff, now you have to take it further and find how much time it takes to hit the bottum.

The relationship that you need to explore is:

Yf = Yi + Vo*t + 1/2*a*t2

This relates position, initial velocity and position to yield time. (The last term will be negative with g in the negative direction.)
 
  • #7
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Ok thanks that equation really helps. I've gotten the first two answers and now the last thing I have left is C. I know how to get C (it is 114.97 meters) thanks to some help I got from a friend. But he really just gave me the answer, I was wondering if you guys could explain the equation he used here

Let y be height reached above the cliff.

0 = u^2 - 2gy
y = u^2 / 2g
= 12.0^2 / (2 * 9.81)
= 7.34 m.

is that v^2 = v_0^2 + 2 a \Delta x ?
 
  • #8
LowlyPion
Homework Helper
3,097
5
Ok thanks that equation really helps. I've gotten the first two answers and now the last thing I have left is C. I know how to get C (it is 114.97 meters) thanks to some help I got from a friend. But he really just gave me the answer, I was wondering if you guys could explain the equation he used here

Let y be height reached above the cliff.

0 = u^2 - 2gy
y = u^2 / 2g
= 12.0^2 / (2 * 9.81)
= 7.34 m.

is that v^2 = v_0^2 + 2 a \Delta x ?

Yes. Delta x here is 95.

Even though you are throwing it up initially, it will be as though you threw it down as it passes the level of the top with the same downward velocity you threw it up with.

Alternatively you can use the total height (95 + 7.34) and use 0 as the v from the top. It should yield the same answer.

Your total distance however should only be twice the 7.34 + 95. 114.9 looks not to be consistent.
 
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