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

[Priyadarshini]

Homework Statement

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 .

Homework Equations

T=(usinA)/g, where A=90
s= 1/2at^2 + ut
GPE=mgh
KE=1/2mv^2

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
The answer is incorrect. The correct answer is 3:1.

 

[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.

 

[andrevdh]

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

 

[Priyadarshini]

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.

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.

 

[Priyadarshini]

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

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.

 

[andrevdh]

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

 

[haruspex]

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.

So what sign should be on the 9.81 gravitational acceleration?

 

[andrevdh]

s= (1/2 x 9.81 x 0.816^2) + (16 x 0.816)
= 16.322m

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

 

[Priyadarshini]

So what sign should be on the 9.81 gravitational acceleration?

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

 

[Priyadarshini]

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

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.

 

[Alex Chen]

Homework Statement

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 .

Homework Equations

T=(usinA)/g, where A=90
s= 1/2at^2 + ut
GPE=mgh
KE=1/2mv^2

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
The answer is incorrect. The correct answer is 3:1.

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.

 

[haruspex]

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

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

 

[jbriggs444]

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

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.

 

[PeroK]

Homework Statement

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 .

Homework Equations

T=(usinA)/g, where A=90
s= 1/2at^2 + ut
GPE=mgh
KE=1/2mv^2

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
The answer is incorrect. The correct answer is 3:1.

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

 

[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...

 

[haruspex]

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...

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

 

[William White]

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

yes, it is 3:1

 

[Priyadarshini]

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

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.

 

[Priyadarshini]

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...

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

 

[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.

 

[Priyadarshini]

Homework Statement

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 .

Homework Equations

T=(usinA)/g, where A=90
s= 1/2at^2 + ut
GPE=mgh
KE=1/2mv^2

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
The answer is incorrect. The correct answer is 3:1.

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.

But isn't the speed at t/2 16m/s?

 

[Priyadarshini]

Homework Statement

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 .

Homework Equations

T=(usinA)/g, where A=90
s= 1/2at^2 + ut
GPE=mgh
KE=1/2mv^2

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
The answer is incorrect. The correct answer is 3:1.

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.

But the speed at the given time (t/2)

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 * 08.15 = 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.

Oh, thank you! But why do you take half the initial speed? At t=0, the speed is 16m/s, but at t/2, how does the initial speed half?

 

[William White]

But the speed at the given time (t/2)

Oh, thank you! But why do you take half the initial speed? At t=0, the speed is 16m/s, but at t/2, how does the initial speed half?

look at the SUVAT equation

(1) v = u + at

at the top of the balls flight, v = 0

so

t = u/a = 1.63 s

half this time is 0.815 s

now plug that back into the equation (1)

a*t = 8 which is half the intial speed

 

[William White]

"how does the initial speed half?"

graph it
https://www.desmos.com/calculator/le5uuyyjay

the blue line is the height (y axis) vs time (x axis)

the red line is velocity (y axis) vs time (x axis)

 

[William White]

But isn't the speed at t/2 16m/s?

no, the initial velocity is + 16m/s

it gets thrown up in the air and is subject to acceleration -9.8 m/s^2 (gravity) which slows it down until it comes to a stop (its maximum height). Then it falls back to Earth again, accelerating at - 9.8m/s^2 and lands on the floor at a velocity of - 16 m/s

 

[Priyadarshini]

look at the SUVAT equation

(1) v = u + at

at the top of the balls flight, v = 0

so

t = u/a = 1.63 s

half this time is 0.815 s

now plug that back into the equation (1)

a*t = 8 which is half the intial speed

Ohhh! Thank you! The graph and the calculation made it clearer!

 

[William White]

deleted...accidental reptition

 

[haruspex]

v^2-u^2=2as
v^2=2 x -9.81 x 9.789

What happened to u²?

This problem illustrates why it is much better to keep everything symbolic as long as possible. If you ignore the given numbers (except the "/2") and just use symbols for the variables you can get to the answer purely in algebra, barely any arithmetic required. There are many other benefits too.

 

[haruspex]

you can do it the long way:

William, I know you are keen to be very helpful to the posters, and no doubt they are grateful, but the way the homework forums work is that we do not post complete solutions (unless the poster has already found one). We ask questions, provide hints, point out mistakes. In the long term, we believe this is of greater benefit.

 

[Priyadarshini]

What happened to u²?

This problem illustrates why it is much better to keep everything symbolic as long as possible. If you ignore the given numbers (except the "/2") and just use symbols for the variables you can get to the answer purely in algebra, barely any arithmetic required. There are many other benefits too.

I had assumed it was 0. But I get it now, u=16m/s.
Thanks!

 

[William White]

William, I know you are keen to be very helpful to the posters, and no doubt they are grateful, but the way the homework forums work is that we do not post complete solutions (unless the poster has already found one). We ask questions, provide hints, point out mistakes. In the long term, we believe this is of greater benefit.

yes, I totally understand, but OP was lost after a week of help; and the help was getting more an more abstract.

One piece of helpful advice was "Note that the answer is independent of g, m and u and, in fact, there was no need for any numerical calculations"

Which, is totally useless for somebody that is lost. Why is the answer independent? Why is there no need for calcs? There was no explanation. That was just left there, hanging. Fine if you are au fait with projectile motion - utterly bewildering if you are not. If the OP was already lost, this "smart" help help just made them even more lost!

Surely there is a point where somebody needs to be shown how to do something so they can move forwards? I know that is certainly true for me - especially after a week of frustration trying to figure something out! Isn't that what teachers do?

 

[PeroK]

yes, I totally understand, but OP was lost after a week of help; and the help was getting more an more abstract.

One piece of helpful advice was "Note that the answer is independent of g, m and u and, in fact, there was no need for any numerical calculations"

Which, is totally useless for somebody that is lost. Why is the answer independent? Why is there no need for calcs? There was no explanation. That was just left there, hanging. Fine if you are au fait with projectile motion - utterly bewildering if you are not. If the OP was already lost, this "smart" help help just made them even more lost!

Surely there is a point where somebody needs to be shown how to do something so they can move forwards? I know that is certainly true for me - especially after a week of frustration trying to figure something out! Isn't that what teachers do?

@William White I can't say I'm happy with this unsolicited abuse. If you're unhappy with what someone else has said to you, then leave me it of, mate!