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# acceleration due to gravity watch

1. Hi Guys,
i have a problem understanding this question if anyone can help me would mean a lot thanks.
a parachutist jumps from a plane flaying at a height of 5000 metres. she falls for 30 seconds before opening her parachute. Take the acceleration due to gravity as 9.8m/s^2
a) if we ignore the effect of air resistance, how fast will the parachutist be falling after 30 seconds?
what i did was to calculate the initial velocity using 5000/30 and then put that into the formula for acceleration a=(v-u)/t with having t 32s as it says on the question after 30 seconds and then calculate v from that.
is that right?
2. I don't understand your method, her initial velocity is 0 not 5000/30, and time time is 30 s not 32 s. But you quoted the correct equation, just solve for v.
3. (Original post by mik1a)
I don't understand your method, her initial velocity is 0 not 5000/30, and time time is 30 s not 32 s. But you quoted the correct equation, just solve for v.
why the height has been given tho? do you not use that at all?
4. (Original post by Alen.m)
why the height has been given tho? do you not use that at all?
It's likely that you'll use the height in the second part of the question, but for finding her velocity at t=30, you do not use the height of the plane.
5. (Original post by aoxa)
It's likely that you'll use the height in the second part of the question, but for finding her velocity at t=30, you do not use the height of the plane.
The fact that it says find her velocity after 30 seconds confuses me still . That's why i took t= 32 seconds because it says after 30 seconds
6. Acceleration is the rate at which velocity changes. So if you have the acceleration of a thing and the amount of time that the thing has been accelerating, you can easily work out its final velocity without needing any other variables.
7. (Original post by Alen.m)
The fact that it says find her velocity after 30 seconds confuses me still . That's why i took t= 32 seconds because it says after 30 seconds
When it asks for 'after 30 seconds' it means when the time reaches 30 seconds. If you find it for any other time you will get a wrong value because the velocity changes as time goes on.

Just rearrange the equation to then solve for v
8. (Original post by Alen.m)
The fact that it says find her velocity after 30 seconds confuses me still . That's why i took t= 32 seconds because it says after 30 seconds
No, at 30 seconds she opens her parachute. 'After 30 seconds' refers to the moment after this happens, not 31/32 seconds but 30.00000000001 but that .000000001 is inconsequential, so you take time as 30 seconds, as quoted in the question.
9. (Original post by the_malis)
When it asks for 'after 30 seconds' it means when the time reaches 30 seconds. If you find it for any other time you will get a wrong value because the velocity changes as time goes on.

Just rearrange the equation to then solve for v
Im guessing that her inital velocity should be the velocity of the plane in which she is jumped off from but i dont think that would be neccessary as we woking on GCSE level . One more question tho, i've sketched a graph of velocity against time for the parachutist but im not sure if im on the right track. Any oponion on it guys? By the way the distance between AB is when she reaches terminal velocity before she opens the parachute and CD is for when she opened the parachute so she slows down for a bit and then reaches her terminal velocity again .
10. (Original post by Alen.m)

Im guessing that her inital velocity should be the velocity of the plane in which she is jumped off from but i dont think that would be neccessary as we woking on GCSE level . One more question tho, i've sketched a graph of velocity against time for the parachutist but im not sure if im on the right track. Any oponion on it guys? By the way the distance between AB is when she reaches terminal velocity before she opens the parachute and CD is for when she opened the parachute so she slows down for a bit and then reaches her terminal velocity again .
In this case, the vertical initial velocity = 0 when they jump out the plane, unless the question tells you otherwise.
11. (Original post by the_malis)
In this case, the vertical initial velocity = 0 when they jump out the plane, unless the question tells you otherwise.
Thanks. Can you please make a comment on my graph as well? Is that suits the question?
12. The general shape is correct. But in reality there will not be kinks at A and C. The velocity doesn't suddenly stop increasing at A, the rate of increase slows gradually so you'll have a curve. The kink at B is still there because it's caused by opening the parachute. Then for the same reaosn as above, there is no kink at C, but a curved path that approaches the horizontal line.
13. (Original post by mik1a)
The general shape is correct. But in reality there will not be kinks at A and C. The velocity doesn't suddenly stop increasing at A, the rate of increase slows gradually so you'll have a curve. The kink at B is still there because it's caused by opening the parachute. Then for the same reaosn as above, there is no kink at C, but a curved path that approaches the horizontal line.
do you reckon i'll lose mark for that?
14. No idea!
15. (Original post by the_malis)
In this case, the vertical initial velocity = 0 when they jump out the plane, unless the question tells you otherwise.
Do you reckon this graph is correct for the movement of parachutist?
16. (Original post by Alen.m)

Do you reckon this graph is correct for the movement of parachutist?

For AB the graph should be a straight diagonal line from the origin.

As acceleration is constant, if it is a velocity time graph, the gradient should be constant at 9.8m/s2

Think of it logically, as time goes on, the parachutist will be gradually increasing in velocity due to the effect of gravity (our acceleration)
17. (Original post by Alen.m)

Do you reckon this graph is correct for the movement of parachutist?
It should look a bit like this.

This is a general diagram btw, explain how the question is shown basically
Attached Images

18. (Original post by the_malis)
It should look a bit like this.

This is a general diagram btw, explain how the question is shown basically
This is what it's shown in the question.
a parachutist jumps from a plane flaying at a height of 5000 metres. she falls for 30 seconds before opening her parachute. Take the acceleration due to gravity as 9.8m/s^2a) if we ignore the effect of air resistance, how fast will the parachutist be falling after 30 seconds? B) sketch a graph of velocity against time for the movement of parachutists

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