# Question regarding calculating the time it takes for an object to fall

## Homework Statement

Given data: Tom throws a stone from a cliff 20m above the water, the initial velocity is 9m/s.

Calculate the time it takes for the rock to fall down.

## Homework Equations

Now, to my questions. I solved this using the equation:
y = - 1/2 gt^2

But, I found myself pretty confused at the moment.

Can I use any of these equations below to solve the problem?
1. 0 = -1/2 gt^2 + 20
2. 20 = -1/2gt^2
3. 20 = 1/2gt^2
4. -20 = -1/2gt^2
5. -20 = 1/2gt^2

My biggest issue is if i put y as y = 20 does that mean that I'm calculating the time it takes for the stone to do 20 metres?

On the other hand, if I put y as y = 0 like i did in the first equation, am I then calculating the same thing, ''when is the stone in the water''

What is actually the difference between these two?

Thank you!

## Answers and Replies

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There is no difference, besides your reference point.
When you use this:
0 = -1/2 gt^2 + 20
you've set the water as the zero point for the stone.

When you use this:
20 = 1/2gt^2
you've set the water as the 20 meter point for the stone. Regardless, the time will be the same.

As for the negatives, that just changes which direction the stone is moving, but that's all based on where you've set the reference point once again. If you're doing projectile motion, I assume you haven't gotten to energy, but these equations rely on where you set Potential Energy to be zero. But rest assured you're doing it right.

Also, anyone correct me if I'm wrong. I'm no expert.

1 person
Hello there,

Correct me if I'm wrong, but when I'm setting y as y = 20, am I then assuming that my zero point is at the cliff? (My reference point), instead of using the water as my zero point(reference point)?

Thank you for your response.

Best regards,

Yes. And the reason that none of this matters to the math portion is that the stone is still traveling 20 meters. It's great that you want to understand the principles though. Now if only I could understand terminal voltage and DC circuits better...

Of course!

I understand, so to my final question(s)! :D, if I'm putting the reference level at the cliff (zero level), then it's actually -20m to the water, however I suppose that I can just use the absolute value of this (20m), because I'm actually interested in the distance that the stone has to travel, and not the direction?

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Question #2
If I place the zero point at the water instead, and I get that the stone has traveled -3m through calculations, does it just mean that the stone has went 3m deep into the water?

Thank you a lot for your time, I hope someone can help you out with your thread! I haven't really studied that yet I believe so I can't really help you on that part!, I'm sorry :(

Best regards,

Last edited:
Question 1: This is correct in theory. IF you maintain awareness and think about your answers. In projectile motion, I usually use absolute values and just take a second to consider my answer. For instance, if the distance is -20m and I set gravity as 9.8 m/s2, I will get a nonreal answer, because this process would require me to take the square root o a negative.

Question 2: No, when you set the water as your zero point, the distance becomes negative. If the water is zero, then the cliff is the 20 meter point. Although the stone is traveling downward, the distance scale has been set in the opposite direction. So if 0 = 1/2gt^2+20, your 20 meters is positive. When you calculate the -3m, that comes off of the positive 20 meters, meaning the stone is now 17 meters above the water.

No problem. This forum has gotten me out of many tight spots. I only hope to repay the community.

Sincerely,

Question 1: This is correct in theory. IF you maintain awareness and think about your answers. In projectile motion, I usually use absolute values and just take a second to consider my answer. For instance, if the distance is -20m and I set gravity as 9.8 m/s2, I will get a nonreal answer, because this process would require me to take the square root o a negative.

Question 2: No, when you set the water as your zero point, the distance becomes negative. If the water is zero, then the cliff is the 20 meter point. Although the stone is traveling downward, the distance scale has been set in the opposite direction. So if 0 = 1/2gt^2+20, your 20 meters is positive. When you calculate the -3m, that comes off of the positive 20 meters, meaning the stone is now 17 meters above the water.

No problem. This forum has gotten me out of many tight spots. I only hope to repay the community.

Sincerely,
I'm sorry but I didn't really get the second question. ;/

If I determine the water as my zero-level, then that means that the distance to the cliff is 20m. Right? ;p

However what do you mean by that the distance is negative if the zero-level is at the water, isn't the distance above the water positive and below the water negative?

If I determine the positive direction downwards, does that mean that the direction the stone is falling is 20m or -20m if the water is the zero level? I seem quite confused on the magnitudes or is the distance always the same? I don't really get what is being affected when I determine the direction upwards or downwards for example :/

I did sadly not get what you mean that the distance is in the opposite direction either, guess I'm just dumb..

Thank you once again and sorry for taking your time. :/

I'm feeling really confused.. :(
Best regards,

If I determine the water as my zero-level, then that means that the distance to the cliff is 20m. Right? ;p
Yes that's correct. And yes, the distance above the water is positive, but your rock is traveling toward the water. So your scale is set to count from 0 to 20 traveling upward, away from the water, while the rock is traveling downward, toward the water. So if the rock travels -3m, then it must be subtracted from the upward counting 20 meters because your scale is in the opposite direction of motion.

Does that make more sense?

Here's a sports analogy. I don't know if this will help or not. Say you are playing soccer and you're on the goal line, inside the goal. Let's set this as a zero point and the half field line as a positive 20 m point. If you dribble 20 m away from the goal, you've traveled positive distance because you're moving forward from zero and toward a positive end on your scale.

Now let's set the zero point at the half line and the 20 m point at the goal line, but you still start at the goal line. When you dribble the 20 m towards the half line, you're moving away from the 20 m point and toward the zero point - negative distance. This second situation is the same as setting the cliff as the 20 m point and the water as zero and allowing the rock to fall.

Physics is hard man. You're not dumb. Everyone learns differently. Hopefully the analogy helps.

Hello, sorry for my late response. Thank you once again for answering.

Ok, so is it like this:

If I place my zero-level at the water, and the positive direction upwards, that mean that the cliff is at +20m. If I then want to drop something off this cliff, then it is also simply travelling -20m downwards? Is this correct?

If I then define positive direction as down instead, does that mean that the cliff is at -20m, and if I drop something from here then it has to travell +20m?

If I place my zero-level at the cliff which is 20m instead, and assume that the positive direction is down, then it has to travell 20m to end into the water?

If I place my zero-level at the cliff and state that negative direction is down, then I can say that it has to travell -20m?

So it's all about what you put your positive and negative direction as?, if you assume that the positive direction is +50m up from the zero-level, then it's also -50m down to the zero-level from the same point?

Did I get this correct? Thank you once again.

Happy New Year by the way! :)

"If I place my zero-level at the water, and the positive direction upwards, that mean that the cliff is at +20m. If I then want to drop something off this cliff, then it is also simply travelling -20m downwards? Is this correct?"

Yes that's correct.

"If I then define positive direction as down instead, does that mean that the cliff is at -20m, and if I drop something from here then it has to travell +20m?"

Yes. It travels 20m to get to zero.

"If I place my zero-level at the cliff which is 20m instead, and assume that the positive direction is down, then it has to travell 20m to end into the water?"

Yep.

"If I place my zero-level at the cliff and state that negative direction is down, then I can say that it has to travell -20m?"

Absolutely.

"So it's all about what you put your positive and negative direction as?, if you assume that the positive direction is +50m up from the zero-level, then it's also -50m down to the zero-level from the same point?"

Yep. Setting a zero-point and deciding which direction will be positive or negative completely determines your answer.

And once again, not a problem.

Happy New Year!

1 person