Finding the rock's theoretical maximum height.

[Matthew117]

Hello everyone. So I have this one question which I thought was going to be pretty easy, but now I struggle with it. It says:
1. The question
A rock is thrown vertically up in the air with an initial velocity of 0.75ms^-1. What would the rock's maximum theoretical height be?

I am currently in high school and usually there would be one more variable given to make solving easier. We also assume that there's no air resistance, just to make the calculations easier on our level.

Homework Equations

The equations I think are relevant to this problem are:
KE = 1/2mv²
GPE = mgh
3. The Attempt at a Solution
In order to solve this I decided to set the two equations equal to each other. The final KE is going to be equal to GPE if we assume that all energy is transferred (please correct me on that one if I'm wrong).

Assuming that acceleration due to gravity = 10ms^-1

(0.75)²m = 10mh
(9/32)m = 10mh
9/32 = 10h
h = 0.02813m
h = 2.81cm

This height looks very little to me, and I'm 100% sure it's wrong. My teacher wrote that question himself, so maybe he forgot to add one of the variables (it's not the first time something like this happened).
My test is this upcoming Wednesday and this really stresses me out as I feel like the answer is obvious but I can't find it...

Any input would be appreciated! Thanks :)

 

[gneill]

Hi Matthew117, Welcome to Physics Forums.

You've done okay with the problem. You did forget to write the "1/2" on the kinetic energy in the first line of the derivation, but I see that you included it anyways in the "9/32" of the second line. Your result is correct.

This sort of problem you can do symbolically and end up with a handy formula that you can use on other occasions. So:
$\frac{1}{2} m v^2 = m g h$
$h = \frac{v^2}{2 g}$

 

[CWatters]

The reason it doesn't go very high is because 0.75m/s isn't very fast. ... Just 1.7 miles an hour.

 

[Matthew117]

Hi Matthew117, Welcome to Physics Forums.

You've done okay with the problem. You did forget to write the "1/2" on the kinetic energy in the first line of the derivation, but I see that you included it anyways in the "9/32" of the second line. Your result is correct.

This sort of problem you can do symbolically and end up with a handy formula that you can use on other occasions. So:
$\frac{1}{2} m v^2 = m g h$
$h = \frac{v^2}{2 g}$

Thank you so so much! I was thinking about the possible answers and this seemed to be the only logical one (the other one I got was over 100 metres and I don't think anyone can throw rocks that high :P).
Oh yes I did forget to include the 0.5 when I was writing the post, sorry. I will take a note of the equation you wrote for me, thank you.

 

[PeroK]

Thank you so so much! I was thinking about the possible answers and this seemed to be the only logical one (the other one I got was over 100 metres and I don't think anyone can throw rocks that high :P).
Oh yes I did forget to include the 0.5 when I was writing the post, sorry. I will take a note of the equation you wrote for me, thank you.

It's good to be able to do a quick check of results to see whether they make sense. You throw something up at less than $1ms^{-1}$; it takes less than $0.1s$ for gravity to stop it; it's average speed is half what it started with. So, less than $0.5m s^{-1}$ average for less than $0.1s$ means less than $5cm$ height.

That's also a neat way to remember the formula you were given:

$h = v_{average}t = (\frac{v}{2})(\frac{v}{g}) = \frac{v^2}{2g}$