I think this problem is correct: acceleration upward at a frat house

Homework Statement

A student throws a set of keys vertically upward to his fraternity brother, who is in a window a distance h above. The brother's outstretched hands catches the keys on their way up at a time t later. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? (Answer should be in terms of h, t, and g.)

Homework Equations

$\Delta$x = v0t + (1/2)at2
v2 = v02 + 2a$\Delta$x

The Attempt at a Solution

Question 1: when did the act of throwing begin? When the keys were released from the hand, or during the wind up of the arm/hand, before releasing the keys? The physics tutor in the math lab told me the initial velocity was 0. But that is not true if at the moment of release the keys were already at a higher velocity.

Question 2: what does "[a]nswer should be in terms of h, t, and g" mean? Do they want actual numbers using h, t, and g; or is this a numberless problem?

a) $\Delta$x = v0t + (1/2)at2

v0 = (1/t)($\Delta$x - (1/2)at2)

v0 = (1/t)(h - (1/2)gt2)

v0 = (h/t) - ((gt)/2)

b) v2 = v02 + 2a$\Delta$x

v2 = ((h/t) - (gt)/2)2 + 2gh

v = +(h/t) - (gt)/2 + 2gh and v = -((h/t) - (gt)/2 + 2gh)

Since they keys were thrown upward and caught, v in this case must be the positive answer above.

Thanks for any help!

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Doc Al
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Question 1: when did the act of throwing begin? When the keys were released from the hand, or during the wind up of the arm/hand, before releasing the keys? The physics tutor in the math lab told me the initial velocity was 0. But that is not true if at the moment of release the keys were already at a higher velocity.
You are correct. By "initial velocity" they mean the speed with which they left the hand.

Question 2: what does "[a]nswer should be in terms of h, t, and g" mean? Do they want actual numbers using h, t, and g; or is this a numberless problem?
They want you to express the answer in terms of those variables. You are not given any actual numbers.

a) $\Delta$x = vot + (1/2)at2

v0 = (1/t)($\Delta$x - (1/2)at2)

v0 = (1/t)(h - (1/2)gt2)

v0 = (h/t) - ((gt)/2)
Careful with signs. If up is positive, what's the acceleration?

You are correct. By "initial velocity" they mean the speed with which they left the hand.

Uh oh, thanks. The problem has now become much harder if the keys had a velocity great than 0 when they were released.

Careful with signs. If up is positive, what's the acceleration?
Acceleration is negative? I think that negative acceleration is deceleration, but I am not sure.

I also think positive velocity means the object is moving up or to the left, and negative velocity is the opposite. But I don't know for sure.

I want to think that acceleration and velocity must always have the same sign.

Velocity is L/T and acceleration is (L/T)/T.

T is always positive (there is no such thing as negative time) and L has to either be positive or negative, but not both at the same time. So acceleration and velocity always have to agree in sign (either both are positive or both are negative)?

Doc Al
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Uh oh, thanks. The problem has now become much harder if the keys had a velocity great than 0 when they were released.
Why do you think that? If the keys had a speed of zero when released they wouldn't even leave the hand, much less rise up to a distance h.

Acceleration is negative?
Yes, the acceleration of gravity is downward and thus negative. In terms of g, what would it be?

I think that negative acceleration is deceleration, but I am not sure.
I don't think the term "deceleration" is very useful.

I also think positive velocity means the object is moving up or to the left, and negative velocity is the opposite. But I don't know for sure.
In this problem, take up as positive. (We only have to deal with the vertical.) So a positive velocity means moving upward.

Doc Al
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I want to think that acceleration and velocity must always have the same sign.
That's not so. Consider a ball thrown straight up into the air. It goes up, then comes down. So the direction (and thus sign) of the velocity must change. But did the direction (and thus sign) of the acceleration change?

Why do you think that? If the keys had a speed of zero when released they wouldn't even leave the hand, much less rise up to a distance h.
Because, honestly, physics freaks me out, and is my hardest course!

Yes, the acceleration of gravity is downward and thus negative. In terms of g, what would it be?
So free fall acceleration or acceleration downward due to gravity is positve acceleration, because the velocity is getting higher [EDIT: in absolute value]. But if the object's velocity is getting slower, or going against gravity, it is negative acceleration. So g in this case is actually -g.

I don't think the term "deceleration" is very useful.
Ok, so negative acceleration means the object is slowing down and the velocity is decreasing in absolute value.

In this problem, take up as positive. (We only have to deal with the vertical.) So a positive velocity means moving upward.
Thank you. I am going to try. It might take me some time to formulate the equation though.

Doc Al
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Because, honestly, physics freaks me out, and is my hardest course!
It can be a brain-buster, all right. Stick with it and you'll do fine.

So free fall acceleration or acceleration downward due to gravity is positve acceleration, because the velocity is getting higher. But if the object's velocity is getting slower, or going against gravity, it is negative acceleration. So g in this case is actually -g.
Think this way: Acceleration is a vector (just like velocity is a vector). The acceleration of something in free fall is downward, and our sign convention says that means it is negative. And the symbol "g" is just a positive constant (equal to 9.8 m/s^2), so that means that the acceleration is -g.

Ok, so negative acceleration means the object is slowing down and the velocity is decreasing in absolute value.
No, negative acceleration just means that the acceleration is downwards. (At least in this problem.) Yes, when you throw something up, the speed decreases. But once it begins falling, its speed will increase again. In both cases, using our sign convention, the acceleration is negative.

Thank you. I am going to try. It might take me some time to formulate the equation though.
You already had the problem solved! Just change the sign of the acceleration.

One more thing: I don't know if v (or vf) is actually 0, since the keys were caught, or if v is right before they were caught, which is really weird that the book asked that question, since this allows for infinite answers, as "just before" can be just about anything, like 0.00001 m/s or 0.1 m/s. There are infinitely smaller figures before the final velocity reaches 0.

Doc Al
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One more thing: I don't know if v (or vf) is actually 0, since the keys were caught, or if v is right before they were caught, which is really weird that the book asked that question, since this allows for infinite answers, as "just before" can be just about anything, like 0.00001 m/s or 0.1 ms.
No reason to think the speed of the keys when they are caught will be zero. Think of vf as the speed of the keys after they rise a distance h.

It can be a brain-buster, all right. Stick with it and you'll do fine.
I will! Thanks for your encouragement! Physics and calculus so far are my favorite courses. Human biology is ok, but there is nothing mathematical about it, or at least the professor is dumbing down the course for us and not making us do the biochemistry part with mathematical equations.

No, negative acceleration just means that the acceleration is downwards. (At least in this problem.) Yes, when you throw something up, the speed decreases. But once it begins falling, its speed will increase again. In both cases, using our sign convention, the acceleration is negative.
Ok, this is the part that confuses me. If negative acceleration is downwards, how is an object being thrown upwards negative? The direction is upwards, so it is positive? Or does negative acceleration mean the vector arrow is pointing downwards, due to the force of gravity coming from that way?

You already had the problem solved! Just change the sign of the acceleration.
Ok, thank you. In this case though, I only had to do some basic algebra to get the variables in the right place. It's always setting up the problem that is hard for me.

No reason to think the speed of the keys when they are caught will be zero. Think of vf as the speed of the keys after they rise a distance h.
Ok, that is a good way to look at it.

But if a train in stopped after braking, the final velocity is 0. So shouldn't it be the same for the keys? Or maybe it is because the train is gradually slowing down to 0, whereas for the keys thrown upwards, it is instantaneous stopping?

Ok, I am going to re-do the problem, and this brings up another problem:

a) $\Delta$x = v0t + (1/2)at2

v0 = (1/t)($\Delta$x - (1/2)at2)

v0 = (1/t)(h - (1/2)(-g)t2)

v0 = (h/t) + ((gt)/2)

b) v2 = v02 + 2a$\Delta$x

v2 = ((h/t) + (gt)/2)2 + 2(-g)h

v = +((h/t) + (gt)/2) - 2gh
and
v = -((h/t) + (gt)/2) - 2gh

v = (h/t) + (gt)/2 - 2gh
and
v = -(h/t) - (gt)/2 - 2gh

So which option for v do I chose above for the answer? The first one, since that was the same is part (a)?

Doc Al
Mentor
But if a train in stopped after braking, the final velocity is 0. So shouldn't it be the same for the keys? Or maybe it is because the train is gradually slowing down to 0, whereas for the keys thrown upwards, it is instantaneous stopping?
All that you are modeling is the motion of the keys while they are in free fall, with the only force acting being gravity. Of course, after they are caught the speed is zero. And before they are thrown the speed is zero. But those facts are trivial and not helpful in solving the problem.

Just like when they give you a problem of a ball being dropped from some height and you are asked to find the speed when it hits the ground. Of course they mean just before it hits the ground, not after it smashes into the ground and comes to rest. (That would be a trivial problem to solve--the answer would always be zero!)

In the case of the train, you are again solving a constant acceleration problem. In that case the final speed is zero. No other forces or accelerations are involved.

Doc Al
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Ok, I am going to re-do the problem, and this brings up another problem:

a) $\Delta$x = v0t + (1/2)at2

v0 = (1/t)($\Delta$x - (1/2)at2)

v0 = (1/t)(h - (1/2)(-g)t2)

v0 = (h/t) + ((gt)/2)
Good.

b) v2 = v02 + 2a$\Delta$x
There's a simpler formula to use to solve this problem; one that will not require taking square roots.

v2 = ((h/t) + (gt)/2)2 + 2(-g)h

v = +((h/t) + (gt)/2) - 2gh
and
v = -((h/t) + (gt)/2) - 2gh
Careful! It looks like you are assuming that
c2 = a2 + b → c = a + b
No!

v = (h/t) + (gt)/2 - 2gh
and
v = -(h/t) - (gt)/2 - 2gh

So which option for v do I chose above for the answer? The first one, since that was the same is part (a)?
This method (done correctly) will give you two answers. One for going up and one for going down. You want the one going up.

But use a different formula and you'll solve it quicker! (Hint: A formula involving time.)

Ok, thank you. I think I see it now! With the train, the final velocity of 0 is part of the overall velocity as the train goes from moving to stopping. But with a ball being dropped, the ball hitting the ground is just incidental, as without that object stopping the motion, the velocity would have gone on for longer.

Good.

There's a simpler formula to use to solve this problem; one that will not require taking square roots.

Careful! It looks like you are assuming that
c2 = a2 + b → c = a + b
No!

This method (done correctly) will give you two answers. One for going up and one for going down. You want the one going up.

But use a different formula and you'll solve it quicker! (Hint: A formula involving time.)
Oh, duh. I made another stupid algebra mistake. I guess I was too worried about the problem to actually think!

Here it is, using a different formula

b) v = v0 + at2

v = (h/t) + (gt)/2 + (-g)t2

v = (h/t) + (gt)/2 - gt2

Doc Al
Mentor
Ok, thank you. I think I see it now! With the train, the final velocity of 0 is part of the overall velocity as the train goes from moving to stopping. But with a ball being dropped, the ball hitting the ground is just incidental, as without that object stopping the motion, the velocity would have gone on for longer.
Exactly!

Doc Al
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Here it is, using a different formula

b) v = v0 + at2
Almost. I think you mean:
v = v0 + at

Oh, I didn't even realize there was a page 2 of this thread. I kept checking back and saw my post at the bottom of page 1, so I thought there were no new updates.

Thank you for catching my mistake Doc Al. I have been so sloppy with physics.

Here it is again, corrected:

b) v = v0 + at

v = (h/t) + (gt)/2 + (-g)t

v = (h/t) + (gt)/2 - gt

I don't know what I would do without this help forum!

Doc Al
Mentor
Good!