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Free Fall Equations

  • Thread starter pringless
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  • #1
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Can anyone help me out with some free fall equations?

and:

Assume: g = 9.8 m/s

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.1 seconds later with an initial speed of 41.16 m/s. They hit the ground at the same time.
How long does it take the first stone to hit the ground? Answer in units of seconds.

How high is the cliff in meters.
 

Answers and Replies

  • #2
Originally posted by pringless
Can anyone help me out with some free fall equations?

and:

Assume: g = 9.8 m/s

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.1 seconds later with an initial speed of 41.16 m/s. They hit the ground at the same time.
How long does it take the first stone to hit the ground? Answer in units of seconds.

How high is the cliff in meters.
Ok, I´ll give it a shot.

They both fall from the exact same height:
h = s

The expression for each stone is as follows:
2s = at^2 //First stone
2s = 2v0 + a(t - 2.1)^2 //Second stone, v0 is the initial speed.

Obviously both expressions must be equal.
at^2 = 2v0 + a(t - 2.1)^2

From that you should be able to derive an expression for t.
 
  • #3
Since they hit the ground at the same time, then the time it takes rock 1 to hit the ground = the rock it takes the 2nd rock to hit the ground. So, just find the rock time of the 2nd rock.


v = v0t - 1/2gt^2

0 = t(v0 - 1/2gt)

0 = 41t - g/2t^2.


solve for t. and you get it. notice there are 2 answers for t, 0 and some other value. The other value is the one you want :p
 

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