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Mechanics help!!!

On a theme park, the carriage is fired vertically downwards from a heigh of 20m above ground level. The carriage is modelled as a particle and moves freely under gravity before the brakes are applied, 10m below ground level. When the brakes are applied, the carriage is travelling at 25ms-1.a) Find the speed of projection of the carriageb) Find the time taken for the carriage to reach ground levelI've been doing this question for ages and I have no clue on what to do, any help is appreciated
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mqb2766
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Original post by jungkook123
On a theme park, the carriage is fired vertically downwards from a heigh of 20m above ground level. The carriage is modelled as a particle and moves freely under gravity before the brakes are applied, 10m below ground level. When the brakes are applied, the carriage is travelling at 25ms-1.a) Find the speed of projection of the carriageb) Find the time taken for the carriage to reach ground levelI've been doing this question for ages and I have no clue on what to do, any help is appreciated

You must have done suvat?
Draw a diagram of the information in the problem and list which suvat variables are known and which you need to find. The suvat equation that you use should then link those variables.
Original post by jungkook123
On a theme park, the carriage is fired vertically downwards from a heigh of 20m above ground level. The carriage is modelled as a particle and moves freely under gravity before the brakes are applied, 10m below ground level. When the brakes are applied, the carriage is travelling at 25ms-1.a) Find the speed of projection of the carriageb) Find the time taken for the carriage to reach ground levelI've been doing this question for ages and I have no clue on what to do, any help is appreciated

Divide the question into two parts, the first 10, and the second 10m. Start with the first 10m and make your suvat list. Use the respective suvat formula to find u. do the same thing for the second 10m to find t.
Original post by am3ricana
Divide the question into two parts, the first 10, and the second 10m. Start with the first 10m and make your suvat list. Use the respective suvat formula to find u. do the same thing for the second 10m to find t.

It's +20m and -10m.
Original post by jungkook123
On a theme park, the carriage is fired vertically downwards from a heigh of 20m above ground level. The carriage is modelled as a particle and moves freely under gravity before the brakes are applied, 10m below ground level. When the brakes are applied, the carriage is travelling at 25ms-1.a) Find the speed of projection of the carriageb) Find the time taken for the carriage to reach ground levelI've been doing this question for ages and I have no clue on what to do, any help is appreciated

a: You have v, s and a, and want u. Which SUVAT equation works for that?
b: You now have u, a and s, and want t. Which SUVAT equation works for that?

Be careful to apply your convention for positive direction consistently.

You will get two solutions for a, but know that it is projected downwards, so can discount one.
(edited 3 years ago)
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sonia05
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Original post by RogerOxon
a: You have v, s and a, and want u. Which SUVAT equation works for that?
b: You now have u, a and s, and want t. Which SUVAT equation works for that?

Be careful to apply your convention for positive direction consistently.

You will get two solutions for a, but know that it is projected downwards, so can discount one.

what would v be in question a? 0?
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sonia05
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Original post by mqb2766
You must have done suvat?
Draw a diagram of the information in the problem and list which suvat variables are known and which you need to find. The suvat equation that you use should then link those variables.

I did do that but I keep getting the wrong answer
Original post by jungkook123
what would v be in question a? 0?

For a:

v is the final speed, at s=30m (+20 to -10), and is 25m/s. a is the acceleration due to gravity, which is ~10m/s^2 (I don't know what you were told to take this as). I'm taking down as the positive direction, with s=0 being the initial point, so that all my numbers should be positive.

Hint: For a, use v2=u2+2asv^2=u^2+2as and solve for uu.
Original post by jungkook123
I did do that but I keep getting the wrong answer

Please post your working.
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Original post by RogerOxon
For a:

v is the final speed, at s=30m (+20 to -10), and is 25m/s. a is the acceleration due to gravity, which is ~10m/s^2 (I don't know what you were told to take this as). I'm taking down as the positive direction, with s=0 being the initial point, so that all my numbers should be positive.

Hint: For a, use v2=u2+2asv^2=u^2+2as and solve for uu.

I finally got the answer, thank you sm!!!!!
Original post by jungkook123
I finally got the answer, thank you sm!!!!!

Good. Did you see that you get two answers for u? Do you understand why?

Have you done part b?
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sonia05
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Original post by RogerOxon
Good. Did you see that you get two answers for u? Do you understand why?

Have you done part b?

no I only got one answer for u which was 6.08, I haven't done part b yet
Original post by jungkook123
no I only got one answer for u which was 6.08, I haven't done part b yet

Your answer is correct, so I'll post the working:

v2=u2−2asv^2=u^2-2as

∴u2=v2−2as=252−2∗9.80665∗30=36.601\therefore u^2=v^2-2as={25}^2-2*9.80665*30=36.601

[text]\therefore u=\pm 6.05 m{s}^{-1} (3 s.f.)

We're told that the carriage is projected downwards, so can ignore the negative result. It's useful to understand it though. If you throw something upwards, with only gravity acting upon it, it later will pass through the initial point with the same speed, but in the other direction. The negative result here is telling you that, if you projected the carriage upwards at the same speed, it would later pass the initial point at the same speed downwards, so either initial speed is a solution to the equation.
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sonia05
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Original post by RogerOxon
Your answer is correct, so I'll post the working:

v2=u2−2asv^2=u^2-2as

∴u2=v2−2as=252−2∗9.80665∗30=36.601\therefore u^2=v^2-2as={25}^2-2*9.80665*30=36.601

[text]\therefore u=\pm 6.05 m{s}^{-1}
(3 s.f.)

We're told that the carriage is projected downwards, so can ignore the negative result. It's useful to understand it though. If you throw something upwards, with only gravity acting upon it, it later will pass through the initial point with the same speed, but in the other direction. The negative result here is telling you that, if you projected the carriage upwards at the same speed, it would later pass the initial point at the same speed downwards, so either initial speed is a solution to the equation.
you were a big help, thank u sm again
Original post by jungkook123
I haven't done part b yet

We now have the initial speed, u, from part a. We know s (distance from the initial point to the ground) and a. We don't know v (the speed at ground level), You need to choose a SUVAT equation that has only s, u, a and t, then solve for t.

Again, you will get two solutions, of which one will be negative. If you draw a graph of displacement verses time, you will see why. For the maths t=0 isn't some special value, but it is practically, as the carriage wasn't moving before it.
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Original post by RogerOxon
We now have the initial speed, u, from part a. We know s (distance from the initial point to the ground) and a. We don't know v (the speed at ground level), You need to choose a SUVAT equation that has only s, u, a and t, then solve for t.

Again, you will get two solutions, of which one will be negative. If you draw a graph of displacement verses time, you will see why. For the maths t=0 isn't some special value, but it is practically, as the carriage wasn't moving before it.

I got the answer t=1.49(3.sf) and t=-2.73(3.sf), considering time can't be negative I got t=1.49s(3.sf)?
Original post by jungkook123
I got the answer t=1.49(3.sf) and t=-2.73(3.sf), considering time can't be negative I got t=1.49s(3.sf)?

I haven't done part b. A quick check is to use v=u+atv=u+at, which gives ~21 m/s. Now use s=u+v2ts=\frac{u+v}{2}t. That gives s as ~20m, which matches, confirming your answer. It's always good to check your answers, especially when it can be done quickly.
(edited 3 years ago)

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