Motion of a body under acceleration of gravity

AI Thread Summary
The discussion revolves around solving a mechanics problem involving a stone thrown vertically upwards, where the goal is to find the height of a ledge based on the times it passes the ledge. Participants explore using the kinematic equation s = ut - 1/2 gt^2, attempting to eliminate the initial velocity u to express height s in terms of gravitational acceleration g and the times t1 and t2. There is confusion about the necessity of t2 and whether it can be eliminated, with suggestions that knowing any two of u, t1, or t2 allows for the calculation of the third. The consensus leans towards simplifying the equation to find s directly, indicating that the problem may not require an actual numerical solution. Ultimately, the discussion emphasizes the importance of understanding the relationships between the variables involved in the motion under gravity.
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I have been looking at this question from a Mechanics Chapter:-

A stone is thrown vertically upwards with a velocity of u metres per second. It passes a ledge in t1 seconds and repasses it t2 seconds after the start. Find the height of the ledge?

Take g = 9.8 metres per second sq

This is the question in entirety.

I have tried using

s = ut + 1/2 a t^2 substituting g in as

s = ut - 1/2 g t^2

If the height is the same at both these times then I have assumed that:-

ut1 -1/2 g t1^2 = ut2 - 1/2 g t2^2

Then factoring out t1 and t2

t1(u - 1/2 g t1) = t2(u - 1/2 g t2)

then failed to see how this could find s!

Then tried rearranging the original formula in terms of u and t1, so:

u = (s + 1/2 g t1^2)/t1

and substituting into

s = ut2 - 1/2 g t2^2

but couldn't proceed.

Can somebody help? I am assuming that I need to find s in terms of g, t1 and t2 rather than an actual value.
 
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That's REALLY the whole question? Its a bit mysterious what they are really after. Do you really want to just eliminate u from the expression for s? That seems to be what you are trying to do.
 
Why can't you just use the formula
d = Vo * t + 1/2 a * t^2
where
d = height
Vo = initial speed = u
t = time = t1
a = 9.8
t = t1
I don't think you can find out the actual value. This would be how you would calculate d using u and t1. Wouldn't this be enough (as I am assuming I actually know the value of the speed and of the first time - they are given in the problem)?
 
That's why I'm groping for what the problem is actually asking.
 
That honestly is the whole question word for word. Unfortunatly I don't even have the answer to work backwards from either.

Re: Dick - Yes that was what I was trying do - to simplify S in terms of g and t1 or t2. Then I was wondering if I could even eliminate u, as I initially assumed that the initial velocity for both equations (s for t1 and s for t2) is the same but thinking about it, this might not be true.

Or maybe find t1 in terms of t2 (or vice versa) and then substitute into...but not sure what!

Re: husky88 - I agree this could potentially be the answer, just that I thought that in a question like this, details like the time t2 arent provided unless they are intended to be used in the final answer.
 
You can eliminate anyone of u,t1,t2. One of them is redundant. E.g. What's the velocity at time (t1+t2)/2? So u/g=(t1+t2)/2. That's all three in one simple equation. Know any two and you know the third. I'm tempted to just write down s=u*t1-g*t1^2/2 and call it a day.
 
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