Free Fall - Calculating Origin/Release

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The discussion revolves around calculating the release point of a dropped ball using stroboscopic photographs that show its movement. The ball travels 2.0 meters in 4 seconds with an acceleration of 0.01 m/s². Participants explore using the equation of motion to solve for the initial position (Xi) and discuss the need for knowing the velocity at specific points to find the distance from the release point to the first measurement point. They clarify that the initial velocity is zero and suggest using average velocities to determine instantaneous velocities at various points. Ultimately, they conclude that the initial velocity can be calculated using the provided parameters, allowing for the determination of the distance from the release point.
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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
 
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Tearsandrille said:

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.
 
Tearsandrille said:
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.
 
Tearsandrille said:
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!
 
Tearsandrille said:
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.
 
  • #10
Tearsandrille said:
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!
 
  • #11
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 )
 

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