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#1
A rock is thrown upwards with a velocity of 6.0 m/s. What is the maximum height it will reach?
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6 years ago
#2
(Original post by pheonixfeather)
A rock is thrown upwards with a velocity of 6.0 m/s. What is the maximum height it will reach?
Do you know the SUVAT equations?
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#3
(Original post by Benniboi1)
Do you know the SUVAT equations?
I dont think so...
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6 years ago
#4
Either use SUVAT or energy considerations.
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6 years ago
#5
(Original post by pheonixfeather)
I dont think so...
You know how much kinetic energy it has, right?

Use the fact that all the Ek will be converted to gravitational potential.
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#6
(Original post by alow)
You know how much kinetic energy it has, right?

Use the fact that all the Ek will be converted to gravitational potential.
But the only numbers it gives you is the velocity and the gravity so how are you supposed to get the kinetic energy?

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6 years ago
#7
(Original post by pheonixfeather)
But the only numbers it gives you is the velocity and the gravity so how are you supposed to get the kinetic energy?

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#8
(Original post by alow)
that doesn't help, I know those equations. Could you show me how you'd do it please?
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6 years ago
#9
(Original post by pheonixfeather)
that doesn't help, I know those equations. Could you show me how you'd do it please?
I would work out how much kinetic energy the particle has in terms of m and then equate that to the gravitational potential at the highest point.
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6 years ago
#10
(Original post by pheonixfeather)
I dont think so...
So the SUVAT equations involve 5 variables: displacement (S), initial velocity (U), final velocity (V), acceleration (A) and time (T). They are:

(1) v = u + at
(2) s = ut + 0.5at^2
(3) v^2 = u^2 +2as
(4) s = 0.5(u + v)t
(5) s= vt - 0.5at^2

You can use the SUVAT equations if you know 2/3 variables and need to find out the others. In your question, we know initial velocity is 6 m/s, we also know acceleration due to gravity on earth is -9.8m/s (the minus sign is important for this example!).

When you throw the rock up and it reaches it's max height before coming down, what is the velocity of the rock at that point?
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6 years ago
#11
(Original post by Benniboi1)
So the SUVAT equations involve 5 variables: displacement (S), initial velocity (U), final velocity (V), acceleration (A) and time (T). They are:

(1) v = u + at
(2) s = ut + 0.5at^2
(3) v^2 = u^2 +2as
(4) s = 0.5(u + v)t
(5) s= vt - 0.5at^2

You can use the SUVAT equations if you know 2/3 variables and need to find out the others. In your question, we know initial velocity is 6 m/s, we also know acceleration due to gravity on earth is -9.8m/s (the minus sign is important for this example!).

When you throw the rock up and it reaches it's max height before coming down, what is the velocity of the rock at that point?
There's no need to use SUVATs, and just stating them without derivation isn't really giving any understanding.
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6 years ago
#12
(Original post by pheonixfeather)
that doesn't help, I know those equations. Could you show me how you'd do it please?
What do you mean, that doesn't help?
You've been given plenty of help but appear to want someone else to do this for you.

In that equation m cancels on both sides.
You have v and g in the question, so find h. (Or delta h as it appears in the one posted)
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6 years ago
#13
(Original post by alow)
There's no need to use SUVATs, and just stating them without derivation isn't really giving any understanding.
Fair point, it's a valid alternate method worth knowing though.
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6 years ago
#14
You have two options. (1/2)mv^2 = mgh so (1/2)v^2 = gh
Rearranging, you get h = v^2/2g so you can work it out that way.

Using suvat, you want s, have u, v and a and don't need t. The equation is v^2 = u^2 + 2as.
Rearranging, you get s = (v^2 - u^2)/2a. As v^2 = 0 this simplifies to exactly what you have above, just with different letters.
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#15
(Original post by dangerworm)
You have two options. (1/2)mv^2 = mgh so (1/2)v^2 = gh
Rearranging, you get h = v^2/2g so you can work it out that way.

Using suvat, you want s, have u, v and a and don't need t. The equation is v^2 = u^2 + 2as.
Rearranging, you get s = (v^2 - u^2)/2a. As v^2 = 0 this simplifies to exactly what you have above, just with different letters.
Aah! I understand now! Thank you!

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#16
(Original post by Stonebridge)
What do you mean, that doesn't help?
You've been given plenty of help but appear to want someone else to do this for you.

In that equation m cancels on both sides.
You have v and g in the question, so find h. (Or delta h as it appears in the one posted)
You seem to misundersand, what I did want was someone to explain it in terms that I would understamd. You seem to be forgetting that not everybody is a genius and can figure out how to do something by information they already know when they were stuck with that knowledge in the first place.

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6 years ago
#17
(Original post by pheonixfeather)
You seem to misundersand, what I did want was someone to explain it in terms that I would understamd. You seem to be forgetting that not everybody is a genius and can figure out how to do something by information they already know when they were stuck with that knowledge in the first place.

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So are you saying you haven't studied the so called SUVAT equations?
If that is the case then you needed to do this via energy considerations.
Your first post stated that you had "no idea" how to start.
So by the time I posted you had been given the energy formula to use, which you then said didn't help.
In what sense did it not help?
I told you to cancel the m and find h from the values left.
What more did you need?
You only needed to rearrange the formula and plug in the numbers.
To claim that we are expecting you to be a genius to do this is to miss the point of this forum and to turn your nose up at the help that was offered.
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6 years ago
#18
Maybe watching a video with some examples may help

There are a few to watch and helped me a lot
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#19
(Original post by Stonebridge)
So are you saying you haven't studied the so called SUVAT equations?
If that is the case then you needed to do this via energy considerations.
Your first post stated that you had "no idea" how to start.
So by the time I posted you had been given the energy formula to use, which you then said didn't help.
In what sense did it not help?
I told you to cancel the m and find h from the values left.
What more did you need?
You only needed to rearrange the formula and plug in the numbers.
To claim that we are expecting you to be a genius to do this is to miss the point of this forum and to turn your nose up at the help that was offered.
Yes, lots of other help was given. No, I do not NEED to study SUVAT equations because I'm only in fourth year. The formula didn't help because I wasn't given a clue as to how to use it. And I'm not able to continue discussing this with you, as I am trying to revise music now. Goodbye.
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6 years ago
#20
(Original post by pheonixfeather)
A rock is thrown upwards with a velocity of 6.0 m/s. What is the maximum height it will reach?
I guess it should be like this-

u= -6 m/s ( taking upward motion as negative)
v= 0 m/s (as the rock's velocity becomes zero after reaching full height)
g= 9.81 m/s^2
h=?

Using equation,
v^2 = u^2 + 2gh
0 = (-6)^2 + (2 x 9.81) h
19.62 h = -36
so, h = - 1.83 m
As this is a negative value and height or distance simply can't be negative on their own, this means the maximum distance or height reached upward is 1.83m.
I hoe this helps a bit!
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