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MedApplicantttt
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#1
I'm really unsure about this one.
At time t = 0, a small stone is thrown vertically upwards with speed 14.7ms−1 from a point A.
At time t = T seconds, the stone passes through A, moving downwards.
The stone is modelled as a particle moving freely under gravity throughout its motion.
Using the model,
(a) find the value of T,
(2)
(b) find the total distance travelled by the stone in the first 4 seconds of its motion.
(4)
(c) State one refinement that could be made to the model, apart from air resistance, that
would make the model more realistic.
(1)
At time t = 0, a small stone is thrown vertically upwards with speed 14.7ms−1 from a point A.
At time t = T seconds, the stone passes through A, moving downwards.
The stone is modelled as a particle moving freely under gravity throughout its motion.
Using the model,
(a) find the value of T,
(2)
(b) find the total distance travelled by the stone in the first 4 seconds of its motion.
(4)
(c) State one refinement that could be made to the model, apart from air resistance, that
would make the model more realistic.
(1)
Last edited by MedApplicantttt; 1 month ago
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MedApplicantttt
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#2
mqb2766
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#3
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#3
(Original post by MedApplicantttt)
For part a I got 1.5s but the answer says 3s?
For part a I got 1.5s but the answer says 3s?
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zoe2404
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#4
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#4
I'm really confused with this too. The mark scheme combines the journey up to the highest point and back down using the equation 14.7 = -14.7 + 9.8T, I didn't even know you could combine the journey up and the journey down ? and why is it -14.7. ahhh
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mqb2766
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#5
(Original post by zoe2404)
I'm really confused with this too. The mark scheme combines the journey up to the highest point and back down using the equation 14.7 = -14.7 + 9.8T, I didn't even know you could combine the journey up and the journey down ? and why is it -14.7. ahhh
I'm really confused with this too. The mark scheme combines the journey up to the highest point and back down using the equation 14.7 = -14.7 + 9.8T, I didn't even know you could combine the journey up and the journey down ? and why is it -14.7. ahhh
v^2 = u^2 + 2*g*0
as the displacement is zero so v^2=u^2 so v=-u. But the projectile motion is a quadratic which is symmetrical about the highest point, so the speeds will be the same.
The force on the body due to gravity (mg) does not change if youre going up or going down so its the same constant acceleration suvat scenario.
Last edited by mqb2766; 1 month ago
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e^iπ + 1 = 0
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#6
(Original post by zoe2404)
I'm really confused with this too. The mark scheme combines the journey up to the highest point and back down using the equation 14.7 = -14.7 + 9.8T, I didn't even know you could combine the journey up and the journey down ? and why is it -14.7. ahhh
I'm really confused with this too. The mark scheme combines the journey up to the highest point and back down using the equation 14.7 = -14.7 + 9.8T, I didn't even know you could combine the journey up and the journey down ? and why is it -14.7. ahhh
u = 14.7, a = 9.8 (as it is accelerating not decelerating) and s = 0 (as point A is at 0 displacement of where it started)
Substitute those values into s = ut + 1/2at^2 and you get a quadratic which will give you the values 0 (as that is when it was first thrown) and 3.
For part b, collect the values again, and use the same equation.
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mqb2766
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#7
(Original post by e^iπ + 1 = 0)
The way I did it was collecting all of the values, so for part a:
u = 14.7, a = 9.8 (as it is accelerating not decelerating) and s = 0 (as point A is at 0 displacement of where it started)
Substitute those values into s = ut + 1/2at^2 and you get a quadratic which will give you the values 0 (as that is when it was first thrown) and 3.
For part b, collect the values again, and use the same equation.
The way I did it was collecting all of the values, so for part a:
u = 14.7, a = 9.8 (as it is accelerating not decelerating) and s = 0 (as point A is at 0 displacement of where it started)
Substitute those values into s = ut + 1/2at^2 and you get a quadratic which will give you the values 0 (as that is when it was first thrown) and 3.
For part b, collect the values again, and use the same equation.
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MedApplicantttt
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#8
Thanks so much.
I'm so confused about part b. Could someone please help. I started to work out the maximum height which I got to be 11m?
I'm so confused about part b. Could someone please help. I started to work out the maximum height which I got to be 11m?
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mqb2766
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#9
(Original post by MedApplicantttt)
Thanks so much.
I'm so confused about part b. Could someone please help. I started to work out the maximum height which I got to be 11m?
Thanks so much.
I'm so confused about part b. Could someone please help. I started to work out the maximum height which I got to be 11m?
You could consider what happens in the next 2.5s from the peak point and then add the two distances together. Or calculate the (positive) displacement from A at 4s, then add the up and down again distances to the peak.
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MedApplicantttt
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#10
(Original post by e^iπ + 1 = 0)
The way I did it was collecting all of the values, so for part a:
u = 14.7, a = 9.8 (as it is accelerating not decelerating) and s = 0 (as point A is at 0 displacement of where it started)
Substitute those values into s = ut + 1/2at^2 and you get a quadratic which will give you the values 0 (as that is when it was first thrown) and 3.
For part b, collect the values again, and use the same equation.
The way I did it was collecting all of the values, so for part a:
u = 14.7, a = 9.8 (as it is accelerating not decelerating) and s = 0 (as point A is at 0 displacement of where it started)
Substitute those values into s = ut + 1/2at^2 and you get a quadratic which will give you the values 0 (as that is when it was first thrown) and 3.
For part b, collect the values again, and use the same equation.
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e^iπ + 1 = 0
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#11
(Original post by MedApplicantttt)
Hi, could you pls help with b
Hi, could you pls help with b
Use a suvat equation with those terms to find the displacement below point A, so it will come out as a negative.
Then work out the distance from point A to the highest point; so v = 0. Find the displacement from that and double it (as it is the displacement for A to the highest point on one side) , and add on your previous displacement (not with the negative sign). Let me know if that matches the mark scheme.
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Spirereyli
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#12
I got an answer of -3, Did I do anything wrong?Btw do you have the mark scheme, I would love to check my answers!!Kind Regards
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sampucket1234
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mqb2766
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#14
(Original post by Spirereyli)
I got an answer of -3, Did I do anything wrong?Btw do you have the mark scheme, I would love to check my answers!!Kind Regards
I got an answer of -3, Did I do anything wrong?Btw do you have the mark scheme, I would love to check my answers!!Kind Regards
Last edited by mqb2766; 1 month ago
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