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Old Sep 26th 2012, 04:26 PM   #1
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Another mechanics problem

1. An astronaut drops a rock off the cliff edge, rock hits the bottom @ 4.15s. 2. He tosses another rock upwards @ a height of 2m and it takes 6.3s for it to fall. Acceleration is unknown. I need to find the height of the cliff.

I tried setting up the equation so that the free fall distance is y=(-1/2)gt^2. How would I go about solving this problem? Do I find the time it takes to reach maximum height, and then just use the free-fall equation to find the distance from the maximum height to the ground - 2m? I tried that and ended up getting a height of 0.52m.. What am I doing wrong?
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Old Sep 27th 2012, 07:06 AM   #2
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Please clarify something - when you say "he tosses another rock upwards @ a height of 2m and it takes 6.3s for it to fall" do you mean the rock reaches an altitude of 2 meters before falling and it takes 6.3s total from the time it leaves his hand to when it hits the ground? Or do you mean that it takes 6.3s to fall from a height that is 2m higher than where the first rock was dropped from?

Assuming the former - you have 3 equations in 3 unknons:

1. From the time it takes to fall from the cliff: 0 = H - (1/2)g(4.15s)^2. where H and g are unknown.

2. From the fact that on the second trial it reaches a max altitude of 2m:
2m = v_0T_2 - (1/2)gT_2^2, where v_0 and g are unknowns, and T_2 = half the additional time, or (1/2)(6.3s-4.15s).

3. From the fact that the total flight time in trial 2 is 6.3s:
0= H + v_0(6.3s) - (1/2)g(6.3s)^2

From these 3 equations you should be able to solve for H, v_0, and g.
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Old Sep 27th 2012, 11:59 AM   #3
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I would disagree with the second equation, though I found that it gives a solution quite close to the one I got using another equation.

And that other equation is:

v^2 = u^2 - 2as

T_2 from your second equation is not an exact time.

It's only the red part that takes 4.15s, and the remaining black path for the stone that was thrown up takes the difference in time.

The time to go up is not necessarily the same it takes for the other black path near to the bottom of the cliff.

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Old Sep 27th 2012, 12:48 PM   #4
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Originally Posted by Unknown008 View Post

T_2 from your second equation is not an exact time.

It's only the red part that takes 4.15s, and the remaining black path for the stone that was thrown up takes the difference in time.

That's what I asked the OP to clarify. The method I described assumes that the 6.3s is the total time, including rising 2m and falling all the way to the ground.
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Old Sep 28th 2012, 08:50 AM   #5
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If it takes 4.15s to fall along h m from rest, then it would take 4.15s to fall along the same distance from rest irrespective of where the rock starts falling.

I have drawn 2 h in the picture I posted (they are of the same length).


If you want some calculations:
h = ut + 0.5at^2

Since the rock is dropped in the first instance, u = 0.
h = 0.5 at^2

In this example, t = 4.15 s

In the second case, the rock is thrown upwards. At the highest point, it is instantaneously at rest and falls back down. The time from the point it's at rest to the a distance h below that same point is also 4.15s, because:
h = ut_2 + 0.5at_2^2

u = 0 again;

h = 0.5at_2^2

Since the same h and a are being involved, t = t_2; or the time to cover h (in blue) is the same as the time required to cover h (in red and labelled in purple).
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