Stone is dropped from the top of a cliff

  • Thread starter Alice
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In summary, Alice is trying to figure out the height of a cliff by using the equation /\X=X-Xo=VoxT+1/2AxT^2 which applies to objects with a constant acceleration. She knows that the stone was dropped from the top of the cliff with an initial velocity of 0 and hit the ground after 3.66 seconds. With this information, she is able to solve for the height of the cliff. A mentor reminds Alice not to post complete solutions, and another user suggests rewriting the equation for y and plugging in the numbers to solve the problem.
  • #1
Alice
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0
Hey

This seems to be way easy, but I'm just not gettng where to start it. A stone is dropped from the top of a cliff, it is seen to hit the ground below after 3.66 s, how high is the cliff? Anyone know? Thanks.
 
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  • #2
Originally posted by Alice
A stone is dropped from the top of a cliff,

That means that vi=0, and the yi the height of the cliff, which is unknown. Implicit in this is that the acceleration of the stone is that of gravity.

it is seen to hit the ground below after 3.66 s,

That means that Δt=3.66s. Implicit in that is that yf=0, which is ground level.

how high is the cliff? Anyone know?

Can you find an equation that relates those quantities?
 
  • #3
Mentor Edit: Please don't post complete solutions. Thank you.
 
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  • #4
Tom-

I pretty much got that far on my own. I just don't know which formula ot use. The one that looks the closes would be

/\=Delta

/\X=X-Xo=VoxT+1/2AxT^2

and

Vx^2-Vox^2=2Ax/\x

but I'm not sure. I'm having a really hard time figuring out which formula goes where. Thanks.

-Alice
 
  • #5
Originally posted by Alice
/\X=X-Xo=VoxT+1/2AxT^2
This one always works for a constant acceleration. And, of course, that "X" doesn't have to be horizontal.
 
  • #6
Originally posted by turin
This one always works for a constant acceleration. And, of course, that "X" doesn't have to be horizontal.

Right, the equation also holds for "y".

Alice, look at the equation that turin referred to, rewrite it for y (just replace x with y), and verify that you do in fact have all the information to solve the problem. Then, plug in the numbers.
 

1. What is the acceleration of the stone as it falls?

The acceleration of the stone is due to gravity and is constant at 9.8 meters per second squared, regardless of the mass of the stone.

2. How does air resistance affect the stone's fall?

Air resistance, also known as drag, slows down the stone's fall by pushing against it in the opposite direction. It becomes more significant as the stone falls faster.

3. What is the formula for calculating the stone's velocity as it falls?

The formula for calculating the stone's velocity is V = gt, where V is the velocity in meters per second, g is the acceleration due to gravity (9.8 m/s²), and t is the time in seconds.

4. How long does it take for the stone to reach the ground?

The time it takes for the stone to reach the ground depends on the height of the cliff. It can be calculated using the formula t = √(2h/g), where t is the time in seconds, h is the height in meters, and g is the acceleration due to gravity.

5. Does the mass of the stone affect its fall?

The mass of the stone does not affect its fall. As stated before, the acceleration due to gravity is constant regardless of the mass of the object. However, a heavier stone may experience more air resistance, causing it to fall slower than a lighter stone.

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