Motion of a body under acceleration of gravity

In summary, Dick was looking for a way to simplify a equation in terms of g and t1 or t2. He tried rearranging the original formula, but couldn't find the answer. He was then told that one of the variables must be eliminated and he decided on t1 in terms of t2.
  • #1
wanchosen
19
0
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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  • #2
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.
 
  • #3
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)?
 
  • #4
That's why I'm groping for what the problem is actually asking.
 
  • #5
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.
 
  • #6
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.
 

1. What is the acceleration of gravity?

The acceleration of gravity, denoted as g, is a constant value of approximately 9.8 meters per second squared (m/s²). This means that for every second an object falls due to gravity, its velocity increases by 9.8 m/s in the downward direction.

2. What factors affect the motion of a body under acceleration of gravity?

The motion of a body under acceleration of gravity is affected by two main factors: the mass of the object and the strength of the gravitational force. The greater the mass of the object, the more force is required to accelerate it. Similarly, the stronger the gravitational force, the greater the acceleration of the object.

3. How does air resistance impact the motion of a body under acceleration of gravity?

Air resistance, also known as drag, is a force that opposes the motion of an object through air. This force increases as the speed of the object increases. As a result, air resistance can slow down the acceleration of a falling object, causing it to eventually reach a terminal velocity where the force of air resistance equals the force of gravity.

4. What is the difference between free fall and motion under acceleration of gravity?

In free fall, an object is falling under the influence of gravity with no other forces acting on it. This means that the object is accelerating at a constant rate of 9.8 m/s². On the other hand, motion under acceleration of gravity refers to the movement of an object that is being acted upon by other forces, such as air resistance, in addition to gravity.

5. Can the acceleration of gravity change?

No, the acceleration of gravity is a constant value on Earth. However, it can vary slightly depending on the location and altitude on Earth. For example, the acceleration of gravity at the North Pole is slightly higher than at the equator due to differences in the shape and rotation of the Earth.

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