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Vertical movement under gravity

I’m trying to make sense of representational graphs in relation to a stone being thrown upwards.

If a stone is thrown upwards from a platform 1.2m above grown at a velocity of 6ms^-1.

What is the velocity when it hits the ground if g=10ms^-2.


So, as we’re 1.2m above ground we can set our ground level at -1.2m

The ball is thrown upwards which would have a negative acceleration due to the force of gravity. The initial velocity is positive due to it being thrown. When it falls back down though, wouldn’t the velocity increase? As well as the acceleration increasing.

So, s=-1.2m , a=-10 u=6

v²=u²+2as
v²=(6)²+2(-10)(-1.2)
v²=60
v=+-7.75, =+8 or, does the velocity decrease due to air resistance?

Do the answers differ depending on air resistance?

Edit: is it due to the fact that velocity has direction and hence negative velocity were as speed would be positive as it has no direction?

Appreciate any confirmation. Please don’t go too technical, explain simply :smile: Lol
(edited 1 year ago)
Original post by KingRich
I’m trying to make sense of representational graphs in relation to a stone being thrown upwards.

If a stone is thrown upwards from a platform 1.2m above grown at a velocity of 6ms^-1.

What is the velocity when it hits the ground if g=10ms^-2.


So, as we’re 1.2m above ground we can set our ground level at -1.2m

The ball is thrown upwards which would have a negative acceleration due to the force of gravity. The initial velocity is positive due to it being thrown. When it falls back down though, wouldn’t the velocity increase? As well as the acceleration increasing.

So, s=-1.2m , a=-10 u=6

v²=u²+2as
v²=(6)²+2(-10)(-1.2)
v²=60
v=+-7.75, =+8 or, does the velocity decrease due to air resistance?

Do the answers differ depending on air resistance?

Edit: is it due to the fact that velocity has direction and hence negative velocity were as speed would be positive as it has no direction?

Appreciate any confirmation. Please don’t go too technical, explain simply :smile: Lol


Im not going to work this out but using A-level mechanics.
but essentially treat it as 2 SUVAT problems added together 1) calculate the height at which the rock reaches maximum height
2) calculate velocity as it impacts the ground after starting from peak height.

the speed will initially decrease as it reaches its maximum height, it will hit zero then increase as it accelerated back towards the ground. Velocity will be dependent on how you define the vectors direction.

Yes, air resistance/drag would impact speed (air resistance is a factor of air density/drag coefficient /cross section area//velocity).. although generally air resistance will have a very small effect unless speed is high (in reality this will be negligible for such a scenario). This is NOT asked for in a-level mechanics.

their are also other effects like rotation of the rock which would affect fluid pressure acting on it as it travels causing acceleration (again this would be negligible), and NOT considered at A-level.

if velocity is negative or positive, it doesn’t really matter, you choose the local coordinate system, just make a decision and stick with it.
Original post by mnot
Im not going to work this out but using A-level mechanics.
but essentially treat it as 2 SUVAT problems added together 1) calculate the height at which the rock reaches maximum height
2) calculate velocity as it impacts the ground after starting from peak height.



That isn't an A level approach - the whole motion is dealt with as the OP has done.
Original post by KingRich
I’m trying to make sense of representational graphs in relation to a stone being thrown upwards.

If a stone is thrown upwards from a platform 1.2m above grown at a velocity of 6ms^-1.

What is the velocity when it hits the ground if g=10ms^-2.


So, as we’re 1.2m above ground we can set our ground level at -1.2m


At this point, by assigning ground level as minus 1.2m, you've effectively said upwards is positive. Good! Doesn't matter which way round you do it, just as long as you are consistent throughout your working.



The ball is thrown upwards which would have a negative acceleration due to the force of gravity. The initial velocity is positive due to it being thrown. When it falls back down though, wouldn’t the velocity increase? As well as the acceleration increasing.


Acceleration is negative because it's in the opposite direction to upwards which you've assigned as positive.
Initial velocity positive - yep.

Velocity is decreasing throughout the motion. When it hits the ground it's speed will be greater than the initial speed, but the velocity will be less (it will be greater in magnitude (in this case), but it's actual value will be less, in the same sense that -20 is less than 4, for example).

Accleration does not increase - it is constant throughout.


So, s=-1.2m , a=-10 u=6

v²=u²+2as
v²=(6)²+2(-10)(-1.2)
v²=60
v=+-7.75, =+8 or, does the velocity decrease due to air resistance?

Do the answers differ depending on air resistance?


The suvat model ignores air resistance, and assumes acceleration is constant.
You get two values +/- 7.75 m/s. If you were to draw a graph of velocity against time, then the positive 7.75 m/s value would occur at some theoretical time before t=0, and so isn't valid.

Actual velocity is minus 7.75, if you take upwards as positive.


Edit: is it due to the fact that velocity has direction and hence negative velocity were as speed would be positive as it has no direction?

Appreciate any confirmation. Please don’t go too technical, explain simply :smile: Lol


Yes. Spot on.
Reply 4
Original post by ghostwalker
At this point, by assigning ground level as minus 1.2m, you've effectively said upwards is positive. Good! Doesn't matter which way round you do it, just as long as you are consistent throughout your working.



Acceleration is negative because it's in the opposite direction to upwards which you've assigned as positive.
Initial velocity positive - yep.

Velocity is decreasing throughout the motion. When it hits the ground it's speed will be greater than the initial speed, but the velocity will be less (it will be greater in magnitude (in this case), but it's actual value will be less, in the same sense that -20 is less than 4, for example).

Accleration does not increase - it is constant throughout.



The suvat model ignores air resistance, and assumes acceleration is constant.
You get two values +/- 7.75 m/s. If you were to draw a graph of velocity against time, then the positive 7.75 m/s value would occur at some theoretical time before t=0, and so isn't valid.

Actual velocity is minus 7.75, if you take upwards as positive.



Yes. Spot on.

Thank you for you detailed confirmation. I didn’t receive any notifications that anyone had responded, hence my delay in replying.

I should have mentioned it instructed that g=9.8 and I have based my result in reflection to significant figures given as advised.

Sorry, I don’t know how to highlight specific parts and reply accordingly.
You said acceleration is constant throughout. Are you speaking about the magnitude of acceleration that has no direction?

You mentioned about displacement and that it doesn’t matter which way I set it? Can you elaborate please?

I shall be posting a separate question shortly, perhaps you could explain that for me too. Thank you
Original post by KingRich
You said acceleration is constant throughout. Are you speaking about the magnitude of acceleration that has no direction?


No, magnitude and direction of acceleration are both constant throughout for you to be able to use the suvat model.


You mentioned about displacement and that it doesn’t matter which way I set it? Can you elaborate please?


You choose one direction to be positive when you begin. This is vital, since these are vector quantites you're dealing with. Though sometimes you'll do it automatically without thinking as one way is obviously the only "reasonable" way to go. Then the sign of any of the values for displacement, velcocity, and acceleration are all determined from that.

In your original post, with up being positive, then the final displacement is negative, acceleration is negative, initial velocity is positive.

If you'd chosen down as being positive, then final displacement would be positive, acceleration would be positive, and initial velocity would be negative.

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