Finding this GCSE physics Question really hard - please help

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

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

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

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

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

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?!?!

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?!?!

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?!?!

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 v²

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

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

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

Still not sure

And thanks rep given

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 !

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

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

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?!?!

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

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

Splendid, time to close the thread one thinks.

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)