# Free Fall - Calculating Origin/Release

• Tearsandrille
In summary, the homework statement states that the student does not know the release point/origin of a dropped ball but does have a stroboscopic photograph of it moving through their frame of reference. The student is suppose to figure out how far above the first point (point A) the ball was dropped. They know the acceleration is 0.01m/s^2 and are trying to solve for Xi using d equal to 2.0m. However, they cannot because they do not know the velocity at either point. They are trying to find the velocity at B (or anywhere else) by solving for d using the equation of motion and knowing the instantaneous velocity at point B.f

## Homework Statement

In this problem I don't know the release point/origin of a dropped ball. I do however have a stroboscopic photograph of it moving through my frame of reference.
In my frame of reference it moves 2.0m in 4 seconds. Now, I am suppose to figure out how far above the first point (point A) the ball was dropped. I know the acceleration is 0.01m/s^2.

## Homework Equations

d = 1/2*a*t^2 + Vi*t + Xi

## The Attempt at a Solution

Can I just solve out for Xi using d equal to 2.0m?
2.0m - 1/2*0.01m/s^2*4^2 = Xi

## Homework Statement

In this problem I don't know the release point/origin of a dropped ball. I do however have a stroboscopic photograph of it moving through my frame of reference.
In my frame of reference it moves 2.0m in 4 seconds. Now, I am suppose to figure out how far above the first point (point A) the ball was dropped. I know the acceleration is 0.01m/s^2.

## Homework Equations

d = 1/2*a*t^2 + Vi*t + Xi

## The Attempt at a Solution

Can I just solve out for Xi using d equal to 2.0m?
2.0m - 1/2*0.01m/s^2*4^2 = Xi

That's not going to work, as you don't know the velocity at either point.

Can you think of a way to generate two equations using that equation of motion?

I know the instantaneous velocity at two different points, but I am unsure of how that will help me. So, in theory I could plug stuff into Vf^2 = Vi^2 + 2*a*d, but that will only give me the distance between the two velocities. Which I know already. I also know that I can use Vf = Vi + a*t again I don't see how that will help me out.

I know the instantaneous velocity at two different points, but I am unsure of how that will help me. So, in theory I could plug stuff into Vf^2 = Vi^2 + 2*a*d, but that will only give me the distance between the two velocities. Which I know already. I also know that I can use Vf = Vi + a*t again I don't see how that will help me out.

One of us is misunderstanding the problem. How do you know the velocity at the two points? I thought all you knew was that it took 4 seconds to move 2m? You don't know the velocity anywhere except at the start (that's a hint!).

Oh, I'm sorry. I have five points (A, B, C, D, and E). I know that between each point is 1 second. So, I have measured between each point and used the midpoint method to find the instantaneous velocity at B and D. I know the initial velocity is 0.0m/s, since it is a free fall.

Oh, I'm sorry. I have five points (A, B, C, D, and E). I know that between each point is 1 second. So, I have measured between each point and used the midpoint method to find the instantaneous velocity at B and D. I know the initial velocity is 0.0m/s, since it is a free fall.
So, what's the velocity at A?

Wait... I think I might get it. If I know the instantaneous velocity on point B, and I know the instantaneous velocity at release (Vi = 0.0m/s) I can use Vf^2 = Vi^2 + 2*a*d. Then all I have to do is minus the distance from A to B to get the distance from release to A!

Wait... I think I might get it. If I know the instantaneous velocity on point B, and I know the instantaneous velocity at release (Vi = 0.0m/s) I can use Vf^2 = Vi^2 + 2*a*d. Then all I have to do is minus the distance from A to B to get the distance from release to A!

That's right. IF you know the velocity at a point you can find d that way. How are you going to find the velocity at B (or anywhere else)?

I know that the instantaneous velocity at point B is equal to the average velocity from point A to point C (considering each point is 1 second apart). Sincethe average velocity between to points is equal to the instantaneous velocity at the midpoint. But, that is the only way I know how to do it.

I know that the instantaneous velocity at point B is equal to the average velocity from point A to point C (considering each point is 1 second apart). Sincethe average velocity between to points is equal to the instantaneous velocity at the midpoint. But, that is the only way I know how to do it.

Sounds good!

If you have: a (0.01) s (2.0) and t (4.0)
Use: u = ( s - ( ½ * a * t ² ) ) / t
To find the intial velocity, then use as the final velocity (v) from the start in the following:
Use: s = ( v ² ) / ( 2 * a )