# Solving Free Fall Motion Homework: Find Time in Seconds

• B18
In summary, the body falls half its total distance in 1.7 seconds. The initial velocity is equal to the final velocity.
B18

## Homework Statement

If a body travels half its total path in the last 1.70 s of its fall from rest, find the total time of its fall (in seconds).

## Homework Equations

Vf=Vi+at
Xf-Xi=Vit+1/2at^2

## The Attempt at a Solution

Well I can see what's going in in the problem (I think). I know that Final velocity of the first half is equal to initial velocity of the second half. I tried solving for (Xf-Xi) for the first half and second half. I used Vf1=Vi2=-9.8t+16.66

Everytime I did this all of my terms would cancel out and I'm left with 4.9t^2=0. I know that the first half of t isn't 0 seconds. I'm starting to lose my confidence on this problem.. I just don't see where else i can attack this problem

Any help would be really appreciated. Thanks so much.

It's hard to start the problem with velocity, since there really isn't anything in the problem to indicate what the velocity might be at the half-way point. You do know that the distance the object follows is the same for each, so see if that helps. Call the distance for each half $d$ (or whatever you want to call it), then use an equation that involves distance for each half of the fall.

jackarms said:
It's hard to start the problem with velocity, since there really isn't anything in the problem to indicate what the velocity might be at the half-way point. You do know that the distance the object follows is the same for each, so see if that helps. Call the distance for each half $d$ (or whatever you want to call it), then use an equation that involves distance for each half of the fall.

I had attempted this. I used displacement as the d. That's when all my terms were canceling out.

If the total fall time is t, t-1.7 is the time when half of the total distance d is covered. Write up both equations: distance from the initial position to halfway,d/2 in time t-1.7, and the distance from the initial position to d in time t. You can divide the equations and solve for t.

ehild

ehild said:
If the total fall time is t, t-1.7 is the time when half of the total distance d is covered. Write up both equations: distance from the initial position to halfway,d/2 in time t-1.7, and the distance from the initial position to d in time t. You can divide the equations and solve for t. ehild

When you say divide the equations do you mean divide one equation by the other? Or divide each term by 2?

Let the total time be t seconds.So the time before falling half the distance is (t-1.7) seconds.
Before falling half the distance:Find the velocity of the particle at T=(t-1.7)seconds.The velocity at t seconds is?(hint:use v=u+at)

After falling half the distance, the velocity you calculated becomes the initial velocity.

Use s1=v2-u2/ 2a for 'Before falling half the distance"

and s2=v'2-u'2/ 2a for "After falling half the distance"

since the distance is the same, you could equate s1 and s2.Good luck :)

B18 said:
When you say divide the equations do you mean divide one equation by the other? Or divide each term by 2?
I mean divide one equation with the other equation.

ehild

Nanosuit said:
Let the total time be t seconds.So the time before falling half the distance is (t-1.7) seconds.
Before falling half the distance:Find the velocity of the particle at T=(t-1.7)seconds.The velocity at t seconds is?(hint:use v=u+at)

After falling half the distance, the velocity you calculated becomes the initial velocity.

Use s1=v2-u2/ 2a for 'Before falling half the distance"

and s2=v'2-u'2/ 2a for "After falling half the distance"

since the distance is the same, you could equate s1 and s2.Good luck :)

I really appreciate the time you took to help me here. I found both s1 and s2 however, for s2 would the final velocity be 0 because it would have hit the ground and stopped?? Or would the final velocity of the second half be Vf=Vi+at
Then Vf=-a(t-1.7s)+a(1.7)

I ended up getting t^2-16.2t+132.25=0 and using quadratic formula both answers were non real. Hmmm

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B18 said:
I really appreciate the time you took to help me here. I found both s1 and s2 however, for s2 would the final velocity be 0 because it would have hit the ground and stopped??
It only stops because of the impact from the ground, an input that's not represented in thye equation. Therefore for the purposes of the equation the final velocity must be just before impact.
I ended up getting t^2-16.2t+132.25=0 and using quadratic formula both answers were non real. Hmmm
Please post your working. Use symbols throughout, not the numeric values provided. That makes it much easier to follow what you are doing and spot errors.

Your signs are inconsistent. "Vf=Vi+at Then Vf=-a(t-1.7s)+a(1.7)" can't be right. Vf, Vi and a all point in the same direction.

If possible, avoid numbers until the last moment. Perhaps things cancel. I can't distinguish what 16.2 and 132.25 stand for.

If s1 = s2 anyway, I will use s.
I like your second equation. If I use it for the whole fall, I get 2s = gt2/2 where g = 9.81 m/s2. If I use it for the first half, I get s = g(t-1.7)2/2

Two eqns, 2 unknowns.

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B18 said:
I ended up getting t^2-16.2t+132.25=0 and using quadratic formula both answers were non real. Hmmm

You do not need the velocities.

The total distance d is traveled by t time: d=g/2 t2.

The last half is taken in 1.7 s, so the time for the first half is t-1.7:

d/2=g/2(t-1.7)2.

Divide the equations. d and g cancel, and you get a quadratic equation for t, with real roots.

ehild

ehild said:
You do not need the velocities.

The total distance d is traveled by t time: d=g/2 t2.

The last half is taken in 1.7 s, so the time for the first half is t-1.7:

d/2=g/2(t-1.7)2.

Divide the equations. d and g cancel, and you get a quadratic equation for t, with real roots.

ehild
I was told only to use kinematic equations with the velocities.

Ah, we are helping in parallel here, sorry.
But now I'm here, I would like to know what you got for s in post #8.
Reason I ask is that nano uses an equation that is not in your list. If you don't know what it stands for, it's easy to misinterpret, forget the missing brackets and end up with imaginary results for t.
Perhaps then we'll also understand the 16.2 seconds and the 132.25 seconds 2 ...

If you don't like to divide equations (like me), you can also substitute d (or s) from the first into the second.

And, as I guessed in #10, you can divide by g on both sides (this is ok, because g is known and not zero) and also by 4.

The quadratic at the bottom is my result which didn't provide any real answers. I attached a picture of my work. The equations I can use are written at the top of the page.

BvU said:
Ah, we are helping in parallel here, sorry.
But now I'm here, I would like to know what you got for s in post #8.
Reason I ask is that nano uses an equation that is not in your list. If you don't know what it stands for, it's easy to misinterpret, forget the missing brackets and end up with imaginary results for t.
Perhaps then we'll also understand the 16.2 seconds and the 132.25 seconds 2 ...

If you don't like to divide equations (like me), you can also substitute d (or s) from the first into the second.

And, as I guessed in #10, you can divide by g on both sides (this is ok, because g is known and not zero) and also by 4.
We can disregard that attempt I solved an equation wrong and it escalated from there.

I was told only to use kinematic equations with the velocities
Reasonable. So we'can use the two equations you listed, right?

For the first half you can still write
d/2 = g/2(t-1.7)2, because that is your 2nd relationship (with Vi = 0).

And you know that half way the velocity v' is g (t-1.7) because that is your 1st relationship.

For the second half you can now write
d/2 = v' * 1.7 + g/2 1.72 because that is your 2nd relationship (with Vi = v').

pick up d/2 from the first half and pick up v' as well:

g/2(t-1.7)2 = g (t-1.7) * 1.7 + g/2 1.72

multiply left and right with 2/g ( number, no problemo)

(t-1.7)2 = 2 (t-1.7) * 1.7 + 1.72

If possible, avoid numbers until the last moment. Perhaps things cancel

I hate to quote myself, but it's really useful: you get the chance to
• let things cancel, like the g/2
• check dimensions a lot more effectively
• re-do things a lot quicker if you discover a mistake somewhere halfway
• earn more points when there still is a calculation error (physics teachers tend look down on the math somewhat ;-)

personally I like the 1.7 thingy (a bit too much probably, I'm sorry :-)

BvU said:
I hate to quote myself, but it's really useful: you get the chance to
• let things cancel, like the g/2
• check dimensions a lot more effectively
• re-do things a lot quicker if you discover a mistake somewhere halfway
• earn more points when there still is a calculation error (physics teachers tend look down on the math somewhat ;-)

personally I like the 1.7 thingy (a bit too much probably, I'm sorry :-)
In that last attempt I left the acceleration due to gravity out until the end. The other numbers are from expanding (t-1.7) and such. When I get off of work I'll sit down and re try the way you just posted BvU. I'm just not sure why I don't see the relationships between the first and second halves using the kinematic equations.

Basically you need to get rid of d and v'. v' is the final speed of the first half and at the same time the initial speed for the second half. You 1st relevant equation helps you get rid of v' and your second does the same for d. Originally there were three unknowns (t,d,v') so you needed three equations.

ehild's post #11 is actually only making use of your 2nd relationship and doing so twice in a very smart and very economic way (much easier to look at first half and whole instead of at first half and second half, or at whole and second half). He only needs 2 equations because he doesn't have v' in there.

So his approach is both legal and economic. Laziness pays out twice: you have less work and less chance of errors. Some people even confuse smart with lazy :-)

BvU said:
Basically you need to get rid of d and v'. v' is the final speed of the first half and at the same time the initial speed for the second half. You 1st relevant equation helps you get rid of v' and your second does the same for d. Originally there were three unknowns (t,d,v') so you needed three equations.

ehild's post #11 is actually only making use of your 2nd relationship and doing so twice in a very smart and very economic way (much easier to look at first half and whole instead of at first half and second half, or at whole and second half). He only needs 2 equations because he doesn't have v' in there.

So his approach is both legal and economic. Laziness pays out twice: you have less work and less chance of errors. Some people even confuse smart with lazy :-)

Solved it. Thanks to everyone that offered advice.

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B18 said:
I went through your previous post and found the quadratic and multiplied by g/2 I solved it and got t=6.344486373 or t=0.4555136271. The value of 6.344486373 wasn't correct. Am I missing something else here?
Very hard to say without seeing your working. And please, please, keep everything in symbolic form, only plugging in numeric values at the final step.

haruspex said:
Very hard to say without seeing your working. And please, please, keep everything in symbolic form, only plugging in numeric values at the final step.

I solved the problem symbolically. I just had an arithmetic error at the end. I finally figured it out.

The body is dropped from rest, with zero initial velocity, that is vi=0. The displacement is function of t: x(t)-xi =vi+g/2 t2. It is the same function all along its travel. No need to consider the motion in the second half of the path different from that in the first half.

You can choose the displacement or the distance traveled as variable: x(t)-xi =s=g/2 t2.
At time t1 the distance traveled is s1=g/2 t12 and the velocity is v1=gt1. At time t2 the body traveled s2=g/2 t22 and its velocity is v2=gt2.

You can use any of the equations given, they follow from each other. If you choose the one vf2-vi2=2gs, you need also the equation vf=vi+gt. Subbing vf into the previous equation, you get the function s=g/2 t2, as vi=0

ehild

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## 1. How do I find the time in seconds for an object in free fall motion?

In order to find the time in seconds for an object in free fall motion, you will need to use the formula t = √(2h/g), where t is the time in seconds, h is the height of the object in meters, and g is the acceleration due to gravity (9.8 m/s²). Simply plug in the given values and solve for t.

## 2. Do I need to consider air resistance when solving for time in free fall motion?

No, when solving for time in free fall motion, we assume that there is no air resistance acting on the object. This is known as the "ideal" or "perfect" free fall scenario.

## 3. What units should I use for the height and acceleration due to gravity when using the time formula?

The height should be measured in meters (m) and the acceleration due to gravity should be measured in meters per second squared (m/s²). This will ensure that the time is calculated in seconds.

## 4. Can I use the time formula for objects thrown horizontally?

No, the time formula for free fall motion only applies to objects dropped or thrown vertically. For objects thrown horizontally, you will need to use the formula t = d/v, where t is the time in seconds, d is the horizontal distance traveled, and v is the horizontal velocity of the object.

## 5. Is there a way to check my calculation for the time in seconds?

Yes, there are a few ways to check your calculation for the time in seconds. One way is to use a calculator to double check your work. Another way is to use the time formula backwards to calculate the height or distance traveled by the object, and see if it matches with the given values. You can also ask a peer or teacher to review your work and provide feedback.

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