A Ball Is Thrown Vertically Upwards, Find The Ratio of PE to KE

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1. Aug 18, 2015

1. The problem statement, all variables and given/known data
A ball of mass 65 g is thrown vertically upwards from ground level with a speed of 16 m s–1. Air resistance is negligible. The ball takes time t to reach maximum height. For time t 2 after the ball has been thrown, calculate the ratio of the potential energy of ball to the kinetic energy of ball .

2. Relevant equations
T=(usinA)/g, where A=90
s= 1/2at^2 + ut
GPE=mgh
KE=1/2mv^2
3. The attempt at a solution
T=(usinA)/g
= 16/9.81
= 1.63s
so t/2= 0.816
at 0.816s,
s= (1/2 x 9.81 x 0.816^2) + (16 x 0.816)
= 16.322m
GPE=mgh
0.065 x 9.81 x 16.322
KE= 1/2mv^2
1/2 x 0.065 x v^2
v^2-u^2=2as
v^2=2 x 9.81 x 16.322
substituting,
KE= 1/2 x 0.065 x 2 x 9.81 x 16.322
= 0.065 x 9.81 x 16.322
So, GPE:KE
= (0.065 x 9.81 x 16.322) / (0.065 x 9.81 x 16.322)
=1:1

Last edited: Aug 18, 2015
2. Aug 18, 2015

haruspex

Be careful with signs. Are you defining positive as up or down? Either way, make sure you are consistent on that with all accelerations, velocities and distances.

3. Aug 18, 2015

andrevdh

Try to work in symbols right up to the very end.

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4. Aug 18, 2015

All the calculations I did dealt with the ball moving upwards till it reached it's maximum height. So I took the upwards direction to be positive.

5. Aug 18, 2015

But how would I find the ratio between the two? Shouldn't I have divided the two?
Yeah, working in symbols would have made it easier.

6. Aug 18, 2015

andrevdh

So why don't you try that if it is easier?

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7. Aug 18, 2015

haruspex

So what sign should be on the 9.81 gravitational acceleration?

8. Aug 19, 2015

andrevdh

Your error is at this point as haruspex mentions.
I seems to get a ratio of 4:1?

Last edited: Aug 19, 2015
9. Aug 20, 2015

It should be negative? But what should I do after getting s= -16.322m?

10. Aug 20, 2015

4:1? The answer says it is 3:1.
In their working they have got (3/4) / (1/4). So the 4s cancel. I didn't get any of these numbers.

11. Aug 20, 2015

Alex Chen

Here's what I would do. Calculate the speed of the ball at the time given. Use that to find the KE of the ball. Then, find the height. Convert that to PE and find the ratio.

12. Aug 20, 2015

haruspex

I didn't say change both signs. It's only the 9.81 that had the wrong sign. The distance must be positive.

13. Aug 20, 2015

jbriggs444

You should not be negating the sign on s. You should be negating your sign on a and then recalculating s. What do you get for s now?
Edit: didn't see haruspex in there. Please pardon the duplication.

14. Aug 20, 2015

PeroK

Note that the answer is independent of $g$, $m$ and $u$ and, in fact, there was no need for any numerical calculations.

15. Aug 20, 2015

William White

is this a trick question?

The ball takes time t to reach maximum height. For time t 2... (twice the time it takes to reach max height?)

ie. the ball is back on the ground so its PE and KE are both zero....unless its rolling around...

16. Aug 20, 2015

haruspex

If you study the working in the OP you can deduce it was meant to say t/2.

17. Aug 20, 2015

William White

ah! I only read the question..the working looked like too much of a mess

yes, it is 3:1

18. Aug 21, 2015

Okay, then--
s= (1/2 x -9.81 x 0.816^2) + (16 x 0.816)
= 9.789
GPE=mgh
Then, 0.065 x -9.81 x 9.789
KE= 1/2mv^2
v^2-u^2=2as
v^2=2 x -9.81 x 9.789
KE= 1/2 x 2 x -9.81 x 9.789 x 0.065
GPE:KE
(0.065 x -9.81 x 9.789)/ (1/2 x 2 x -9.81 x 9.789 x 0.065)
= 1:1
The answer still doesn't seem to match.

19. Aug 21, 2015

I'm so sorry, it is supposed to be t/2!

20. Aug 21, 2015

William White

you can do it the long way:

total energy at time of ball release is KE

KE = 1/2 mv^2 = 8.32 J

Time to reach maximum height =
t = u / -g = 1.63s

Half this time is 0.815s

v = u + a * 0.815 = 8 (note this, it is half initial speed)

KE at this t - 0.815

1/2 mv^2 = 2.08 J

which is a quarter of the intial KE (note this, it is quarter initial KE)

The total energy is 8.32J
at t/2
KE = 2.08J and PE is the remainder 6.24J

so the ratio is 3:1

The bits on bold allude to the fact that you did not have to calulate anything.

Last edited: Aug 21, 2015