Stuck on M1 vertical motion? Watch

Rascacielos
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According to my notes, I understood this perfectly a couple of months ago, but now I don't have a clue, particularly about whether to take acceleration as positive or negative. I understand that if its decelerating, it's -9.8 but if I'm taking the whole journey into account, where the particle is projected upwards and decelerates, but then changes direction (i.e. falls back to earth) and accelerates, so I never know whether to take a to = 9.8 or -9.8 when considering a whole journey. Obviously I can't put both 9.8 and -9.8 into the same SUVAT eqn. I hope that makes sense.

I'm struggling with this question in particular anyway. As detailed an explanation as possible would be appreciated:

"A ball is thrown vertically upwards with a speed of 29m/s. It hits the ground 6 secs later. By modelling the ball as a particle find the height above the ground from which it was thrown."
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jaroc
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(Original post by Rascacielos)
According to my notes, I understood this perfectly a couple of months ago, but now I don't have a clue, particularly about whether to take acceleration as positive or negative. I understand that if its decelerating, it's -9.8 but if I'm taking the whole journey into account, where the particle is projected upwards and decelerates, but then changes direction (i.e. falls back to earth) and accelerates, so I never know whether to take a to = 9.8 or -9.8 when considering a whole journey. Obviously I can't put both 9.8 and -9.8 into the same SUVAT eqn. I hope that makes sense.
To avoid confusion, always label one direction as positive or negative. It's up to you what you choose. The simplest way would be to label everything which is up as positive, and down as negative alike. As you say, you need to be consistent with your notation. That is, if you choose up as positive, any speed directed upwards is positive, gravitational acceleration is negative, any height above the 0 height is positive etc.

Your question could be a good example to see this.

"A ball is thrown vertically upwards with a speed of 29m/s. It hits the ground 6 secs later. By modelling the ball as a particle find the height above the ground from which it was thrown."
Recall this equation of motion: x(t)=x_0+v_0t+\tfrac{1}{2}at^2.

Now you can rewrite the equation using the information given in the question, taking up as positive. Then 0=H_0+v_0t-\tfrac{1}{2}gt^2. That is the way I prefer to write it - I include the sign of the gravitational acceleration already in the equation. But you can also write the RHS of the equation as H_0+v_0t+\tfrac{1}{2}gt^2 - then you need to remember to substitute a negative value for g.

Spoiler:
Show
Just to show you that it's totally up to you which direction is positive, you can take down as positive. Then, including the signs already in the equation, you would obtain the following equation 0=-H_0-v_0t+\frac{1}{2}gt^2. I wrote -H_0 because it says in the question that the initial height was above the ground.

In both cases, if you obtain a negative value for H_0, then it means that the point from which the ball was thrown was below the ground.


Basically, you need to decide whether you want to include the sign of gravitational acceleration already in the equation - then (taking up as positive) you write -g and substitute 9.8m/s/s for g - or you prefer to include the sign of gravitational acceleration in the numerical calculations - then you write +g and substitute -9.8m/s/s for g.
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Ollie901
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(Original post by Rascacielos)
I'm struggling with this question in particular anyway. As detailed an explanation as possible would be appreciated:

"A ball is thrown vertically upwards with a speed of 29m/s. It hits the ground 6 secs later. By modelling the ball as a particle find the height above the ground from which it was thrown."
I personally would split that question in two to avoid this confusion.

On the way up, v=29, a=-9.8 u=0. Can work out S and T
On the way back down, u=0, a=9.8, t=(6-T from part 1). Find S here.

S from second bit - S from first bit = Height it was thrown from.
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Rascacielos
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(Original post by Ollie901)
I personally would split that question in two to avoid this confusion.

On the way up, v=29, a=-9.8 u=0. Can work out S and T
On the way back down, u=0, a=9.8, t=(6-T from part 1). Find S here.

S from second bit - S from first bit = Height it was thrown from.
Do you mean u is 29 in the first part, not v?
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Rascacielos
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(Original post by jaroc)
To avoid confusion, always label one direction as positive or negative. It's up to you what you choose. The simplest way would be to label everything which is up as positive, and down as negative alike. As you say, you need to be consistent with your notation. That is, if you choose up as positive, any speed directed upwards is positive, gravitational acceleration is negative, any height above the 0 height is positive etc.

Your question could be a good example to see this.



Recall this equation of motion: x(t)=x_0+v_0t+\tfrac{1}{2}at^2.

Now you can rewrite the equation using the information given in the question, taking up as positive. Then 0=H_0+v_0t-\tfrac{1}{2}gt^2. That is the way I prefer to write it - I include the sign of the gravitational acceleration already in the equation. But you can also write the RHS of the equation as H_0+v_0t+\tfrac{1}{2}gt^2 - then you need to remember to substitute a negative value for g.

Spoiler:
Show
Just to show you that it's totally up to you which direction is positive, you can take down as positive. Then, including the signs already in the equation, you would obtain the following equation 0=-H_0-v_0t+\frac{1}{2}gt^2. I wrote -H_0 because it says in the question that the initial height was above the ground.

In both cases, if you obtain a negative value for H_0, then it means that the point from which the ball was thrown was below the ground.


Basically, you need to decide whether you want to include the sign of gravitational acceleration already in the equation - then (taking up as positive) you write -g and substitute 9.8m/s/s for g - or you prefer to include the sign of gravitational acceleration in the numerical calculations - then you write +g and substitute -9.8m/s/s for g.
(Original post by Ollie901)
I personally would split that question in two to avoid this confusion.

On the way up, v=29, a=-9.8 u=0. Can work out S and T
On the way back down, u=0, a=9.8, t=(6-T from part 1). Find S here.

S from second bit - S from first bit = Height it was thrown from.
Got it! Thank you
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Ollie901
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(Original post by Rascacielos)
Do you mean u is 29 in the first part, not v?
Yeah I did. Sorry.
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Rascacielos
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(Original post by Ollie901)
Yeah I did. Sorry.
No worries!
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