What is the correct formula for solving time in free fall?

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In summary: To determine the relationship between the initial and final velocities, we can use the principle of conservation of energy. The ball is thrown up with some initial kinetic energy, which is converted into potential energy at its maximum height. As it falls back down, that potential energy is converted back into kinetic energy. Therefore, at the point of catching the ball, it should have the same kinetic energy as it did when it was thrown up, just in the opposite direction. This means that the final velocity should be equal in magnitude but opposite in direction to the initial velocity. This is why we use the equation v=v0 + at, because we know that the final velocity (v) is equal to the initial velocity (v0) plus the change in velocity due
  • #1
ammora313
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Homework Statement


t=2v/g

the AAMC used this formula to solve a problem that was asking what would happen to time if the gravity becomes g/6. the answer is that the time increases by a factor of 6.

Homework Equations


v=v0 + at

The Attempt at a Solution



I got the wrong answer by thinking that i can use the free fall equation x=1/2 gt2 Thanks in advance.
 

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  • #2
ammora313 said:

Homework Statement


t=2v/g

the AAMC used this formula to solve a problem that was asking what would happen to time if the gravity becomes g/6. the answer is that the time increases by a factor of 6.

Homework Equations


v=v0 + at

The Attempt at a Solution



I got the wrong answer by thinking that i can use the free fall equation x=1/2 gt2 Thanks in advance.
You need to post the entire original question.
 
  • #3
haruspex said:
You need to post the entire original question.
 

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  • #4
Next time, please try to post the image the right way up.
ammora313 said:
thinking that i can use the free fall equation x=1/2 gt2
But you don't know x, so how can you use that? Did you assume the height reached would be the same in both gravities?
There are 5 variables in the 5 SUVAT equations. Each equation uses four of them. Given the values of three and the need to determine a fourth, pick the one with those four.
Which three do you know here, and which is to be determined?
 
  • #5
I'm sorry about that, I kept trying to make it right side up but it wouldn't budge.

The one I need to determine is the time. The ones I know are acceleration (gravity) and velocity.
There is the equation v=v0 + 1/2 at2 but I'm not sure how to manipulate it to come out to t=2v/g. When I tried to make it equal to t, I get t=v-v0/a. I don't understand where they get 2v from since the v0 is added on one side, wouldn't it get subtracted if it moves to the other side?
 
  • #6
ammora313 said:
There is the equation v=v0 + 1/2 at2 but I'm not sure how to manipulate it
There is no such SUVAT equation. Do you mean s= v0 t + (1/2) at2 , or v=v0 + at?
Also, to what process are you applying this? E.g. the whole process from throwing the ball up to its landing again?
 
  • #7
I'm sorry that was a typo. The equation was v=v0 + at
I was applying it to the whole process, from when the ball was thrown up to it landing
 
  • #8
ammora313 said:
I'm sorry that was a typo. The equation was v=v0 + at
I was applying it to the whole process, from when the ball was thrown up to it landing
OK, so v0 is the velocity at which it was thrown up. What's the velocity just before it lands?
 
  • #9
the velocity before it lands is v=√2gh
 
  • #10
ammora313 said:
the velocity before it lands is v=√2gh
OK, you can do it that way if you can figure out h. Alternatively, what do you think the relationship is between the velocity it was thrown up with and the velocity it lands with?
 
  • #11
haruspex said:
OK, you can do it that way if you can figure out h. Alternatively, what do you think the relationship is between the velocity it was thrown up with and the velocity it lands with?
Would the two velocities by equal? 2v?
 
  • #12
ammora313 said:
Would the two velocities by equal?
Not equal, exactly. Remember, a velocity is a vector, it has direction.
ammora313 said:
2v?
What are you asking?
 
  • #13
haruspex said:
Not equal, exactly. Remember, a velocity is a vector, it has direction.

What are you asking?
I'm trying to figure out how they came up with t=2v/g
 
  • #14
ammora313 said:
I'm trying to figure out how they came up with t=2v/g
Yes, I know that, but your last post finished with the question "2v?". I have no idea what you are asking there.
 
  • #15
Oh ok sorry about that, I didn't know if that's where the 2v in the equation came from. If they're talking about the initial v and the v before it lands.
 
  • #16
ammora313 said:
Oh ok sorry about that, I didn't know if that's where the 2v in the equation came from. If they're talking about the initial v and the v before it lands.
OK, let's get back to where we were. What is the relationship between the initial and final velocities? They are not equal because they are not in the same direction, but it is a very simple relationship.
 
  • #17
Are they inversely proportional to each other? Equal but opposite signs?
 
  • #18
ammora313 said:
Are they inversely proportional to each other? Equal but opposite signs?
Two very different propositions. Don't guess - which is it? You should be able to pick the right one from everyday experience.
 
  • #19
Ok they are inversely proportional to each other since they are vectors
 
  • #20
ammora313 said:
Ok they are inversely proportional to each other since they are vectors
Do you understand what inversely proportional means? It would mean that one of them equals some constant divided by the other. There is no such process as division by a vector.
If you throw a ball up at 3m/s and catch it at the same height as it comes down, how fast do you think it will be traveling when you catch it?
 
  • #21
It'll be traveling at 3 m/s when I catch it. I'm sorry I guess I meant they are proportional to each other but that would be wrong since they are equal with just opposite directions
 
  • #22
ammora313 said:
It'll be traveling at 3 m/s when I catch it. I'm sorry I guess I meant they are proportional to each other but that would be wrong since they are equal with just opposite directions
Right, they are in opposite directions, so there is a minus sign that needs to fit in there somewhere.

So plug that into [itex] v_f = v_i + at [/itex] noting that one of the velocities is a negative value (and taking appropriate care in whether the acceleration is up or down [positive or negative], with your chosen conventions).

There's another way to approach this derivation you don't like working with the velocity formula. Start with

[tex] y = vt + \frac{1}{2}a t^2 [/tex]

where [itex] v [/itex] is the initial velocity. You know that the ball finishes at the same location that it starts. In other words, you know that the ball starts at y = 0 and ends at y = 0. Solve for t. (Substitute whatever you need to in for a, keeping attention to whether it is up or down [positive or negative] given your chosen conventions.)
 
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  • #23
Ok so if I let Vi be negative:

vf = -vi + at

If I move vi to the other side it becomes 2v

2v=at
2v/a=t
2v/g=t

is that correct?
 
  • #24
ammora313 said:
Ok so if I let Vi be negative:

vf = -vi + at

If I move vi to the other side it becomes 2v

2v=at
2v/a=t
2v/g=t

is that correct?

That is one valid way to do it, yes. In that convention that you chose, down is positive. Since the ball initially thrown up, you tacked the negative sign on vi (and this assumes that vi itself is a positive number, equal to vf).

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Edit: Or perhaps even better, you could have kept the equation in it's original form, vf = vi + at, and then made the substitution that vf = -vi. That way the formula remains the same, and it's only the values that change sign.
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You also could have done it using the other convention where up is positive and down is negative. In that case a = -g, the initial velocity is positive and the final velocity is negative.

But if you didn't figure out before-hand that |vi| = |vf| then you could have derived the equation with [itex] y_f = y_i + vt + \frac{1}{2}at^2 [/itex] knowing full well that [itex] y_i = y_f [/itex], using whichever convention that you chose (the convention of whether up or down is positive). So that's yet another way.
 
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  • #25
Thanks so much for your help and for being patient with me. You're awesome!
 

FAQ: What is the correct formula for solving time in free fall?

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