# Finding this GCSE physics Question really hard - please help Watch

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A ball of mass 0.5 kg is thrown directly up with a speed of 6ms

(1) Its maximum gain in potential energy

(2) The maximum height gained

^{-1}. Calculate:(1) Its maximum gain in potential energy

(2) The maximum height gained

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#3

(Original post by

A ball of mass 0.5 kgis thrown directly up with a speed of 6ms

(1) Its maximum gain in potential energy

(2) The maximum height gained

**Sam00**)A ball of mass 0.5 kgis thrown directly up with a speed of 6ms

^{-1}. Calculate:(1) Its maximum gain in potential energy

(2) The maximum height gained

suvat is also a possibility for part 2 - you should find it doesn't contradict the result of your conservation of energy calculation

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(Original post by

here you could use conservation of energy... KE of ball at launch = mgh

suvat is also a possibility for part 2 - you should find it doesn't contradict the result of your conservation of energy calculation

**Joinedup**)here you could use conservation of energy... KE of ball at launch = mgh

suvat is also a possibility for part 2 - you should find it doesn't contradict the result of your conservation of energy calculation

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#5

(Original post by

A ball of mass 0.5 kg is thrown directly up with a speed of 6ms

(1) Its maximum gain in potential energy

(2) The maximum height gained

**Sam00**)A ball of mass 0.5 kg is thrown directly up with a speed of 6ms

^{-1}. Calculate:(1) Its maximum gain in potential energy

(2) The maximum height gained

So just work out kinetic energy for part (1)

Second part...well I do A-level Physics & would use SUVAT, if you do that in GCSE do SUVAT at v=0 lol

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#6

(Original post by

Gain in potential energy = loss in Kinetic energy

So just work out kinetic energy for part (1)

Second part...well I do A-level Physics & would use SUVAT, if you do that in GCSE do SUVAT at v=0 lol

**KINGYusuf**)Gain in potential energy = loss in Kinetic energy

So just work out kinetic energy for part (1)

Second part...well I do A-level Physics & would use SUVAT, if you do that in GCSE do SUVAT at v=0 lol

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#7

(Original post by

So do I need to work out part 2 before I can work out part 1?

**Sam00**)So do I need to work out part 2 before I can work out part 1?

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**KINGYusuf**)

Gain in potential energy = loss in Kinetic energy

So just work out kinetic energy for part (1)

Second part...well I do A-level Physics & would use SUVAT, if you do that in GCSE do SUVAT at v=0 lol

So for part 1 the answer is 9J but I need to learn to method used to get the 9J?

This is what I don't understand, however if I knew the height I would be able to work out the first question?!?!

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#9

(Original post by

I have an exam tomorrow and I'm panicking,

So for part 1 the answer is 9J but I need to learn to method used to get the 9J?

This is what I don't understand, however if I knew the height I would be able to work out the first question?!?!

**Sam00**)I have an exam tomorrow and I'm panicking,

So for part 1 the answer is 9J but I need to learn to method used to get the 9J?

This is what I don't understand, however if I knew the height I would be able to work out the first question?!?!

^{1}⁄

_{2}× mass × velocity

^{2}

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#10

**Sam00**)

I have an exam tomorrow and I'm panicking,

So for part 1 the answer is 9J but I need to learn to method used to get the 9J?

This is what I don't understand, however if I knew the height I would be able to work out the first question?!?!

REMEMBER THIS WITH YOUR LIFE & YOU WILL SUCCEED:

LOSS OF GPE = GAIN IN KE

GPE = mgh

KE = 1/2mv^2

so mgh = 1/2mv^2

So they gave you mass & velocity, you can work out the kinetic energy as you know kinetic energy is = to potential energy

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#11

**Sam00**)

I have an exam tomorrow and I'm panicking,

So for part 1 the answer is 9J but I need to learn to method used to get the 9J?

This is what I don't understand, however if I knew the height I would be able to work out the first question?!?!

on the way up it's converting KE to GPE and on the way down it's converting GPE back to KE. The peak value of GPE occurs when KE=0 and KE equals zero when it's velocity is zero.

the maximum efficiency of an energy conversion is 100%... so part 1 is just asking you to find the value of h at which mgh=1/2 m v

^{2}

you already know m,g & v so it shouldn't be too difficult.

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(Original post by

throwing a ball straight up in the air starts an energy conversion process.

on the way up it's converting KE to GPE and on the way down it's converting GPE back to KE. The peak value of GPE occurs when KE=0 and KE equals zero when it's velocity is zero.

the maximum efficiency of an energy conversion is 100%... so part 1 is just asking you to find the value of h at which mgh=1/2 m v

you already know m,g & v so it shouldn't be too difficult.

**Joinedup**)throwing a ball straight up in the air starts an energy conversion process.

on the way up it's converting KE to GPE and on the way down it's converting GPE back to KE. The peak value of GPE occurs when KE=0 and KE equals zero when it's velocity is zero.

the maximum efficiency of an energy conversion is 100%... so part 1 is just asking you to find the value of h at which mgh=1/2 m v

^{2}you already know m,g & v so it shouldn't be too difficult.

I will now attempt the second part

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(Original post by

REMEMBER THIS WITH YOUR LIFE & YOU WILL SUCCEED:

LOSS OF GPE = GAIN IN KE

GPE = mgh

KE = 1/2mv^2

so mgh = 1/2mv^2

So they gave you mass & velocity, you can work out the kinetic energy as you know kinetic energy is = to potential energy

**KINGYusuf**)REMEMBER THIS WITH YOUR LIFE & YOU WILL SUCCEED:

LOSS OF GPE = GAIN IN KE

GPE = mgh

KE = 1/2mv^2

so mgh = 1/2mv^2

So they gave you mass & velocity, you can work out the kinetic energy as you know kinetic energy is = to potential energy

Still not sure

And thanks rep given

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#14

(Original post by

I've worked out part 1 but need to do part 2 using SUVAT?

Still not sure

And thanks rep given

**Sam00**)I've worked out part 1 but need to do part 2 using SUVAT?

Still not sure

And thanks rep given

You can use the LOSS OF POTENTIAL ENERGY = GAIN IN KINETIC ENERGY formula

So mgh = 1/2mv^2

Plot in your values & find h, the height !

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(Original post by

Have you done SUVAT? If you haven't, don't worry.

You can use the LOSS OF POTENTIAL ENERGY = GAIN IN KINETIC ENERGY formula

So mgh = 1/2mv^2

Plot in your values & find h, the height !

**KINGYusuf**)Have you done SUVAT? If you haven't, don't worry.

You can use the LOSS OF POTENTIAL ENERGY = GAIN IN KINETIC ENERGY formula

So mgh = 1/2mv^2

Plot in your values & find h, the height !

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#16

(Original post by

I am sorry I don't follow

**Sam00**)I am sorry I don't follow

You know loss in potential energy = gain in kinetic energy

Since (gravitational) potential energy = mgh

& kinetic energy = 1/2mv^2

you can equate them together since they are equal

so mgh = 1/2mv^2

m = mass

g= gravitational field strength

h = height

v= velocity

So they gave you mass & velocity. Now rearrage mgh = 1/2mv^2 to make the height, h the subject & subtitute the values given

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(Original post by

You know loss in potential energy = gain in kinetic energy

Since (gravitational) potential energy = mgh

& kinetic energy = 1/2mv^2

you can equate them together since they are equal

so mgh = 1/2mv^2

m = mass

g= gravitational field strength

h = height

v= velocity

So they gave you mass & velocity. Now rearrage mgh = 1/2mv^2 to make the height, h the subject & subtitute the values given

**KINGYusuf**)You know loss in potential energy = gain in kinetic energy

Since (gravitational) potential energy = mgh

& kinetic energy = 1/2mv^2

you can equate them together since they are equal

so mgh = 1/2mv^2

m = mass

g= gravitational field strength

h = height

v= velocity

So they gave you mass & velocity. Now rearrage mgh = 1/2mv^2 to make the height, h the subject & subtitute the values given

0.5(0.5 x 6^2) = 9J

h = 9/0.5x9.81

h = 1.83

Thank you kind sir

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#18

**Sam00**)

A ball of mass 0.5 kg is thrown directly up with a speed of 6ms

^{-1}. Calculate:

(1) Its maximum gain in potential energy

(2) The maximum height gained

**Sam00**)

I have an exam tomorrow and I'm panicking,

So for part 1 the answer is 9J but I need to learn to method used to get the 9J?

This is what I don't understand, however if I knew the height I would be able to work out the first question?!?!

(Original post by

ok now the first part makes sense, 0.5(0.5x6^2) = 9J

I will now attempt the second part

**Sam00**)ok now the first part makes sense, 0.5(0.5x6^2) = 9J

I will now attempt the second part

(Original post by

0.5 x 9.81 x h = 9J

0.5(0.5 x 6^2) = 9J

h = 9/0.5x9.81

h = 1.83

Thank you kind sir

**Sam00**)0.5 x 9.81 x h = 9J

0.5(0.5 x 6^2) = 9J

h = 9/0.5x9.81

h = 1.83

Thank you kind sir

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#19

Height is calculated using the GPE equation. Assuming you have calculated gain in potential energy in part 1, rearrange Ep = mgh to height = potential energy / (mass * 9.81)

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#20

(Original post by

Splendid, time to close the thread one thinks.

**04MR17**)Splendid, time to close the thread one thinks.

(Original post by

Height is calculated using the GPE equation. Assuming you have calculated gain in potential energy in part 1, rearrange Ep = mgh to height = potential energy / (mass * 9.81)

**AndrewMarriott3**)Height is calculated using the GPE equation. Assuming you have calculated gain in potential energy in part 1, rearrange Ep = mgh to height = potential energy / (mass * 9.81)

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