# Need help ASAP - a level maths mechanics exam question

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If anyone could help it would be greatly appreciated!

The height of a pole vaulter above the ground can be modelled using the equation h=1/60(125x - 12x^2) , where h metres is the vertical height of the pole vaulter and x metres is the horizontal distance travelled after his feet leave the ground.

a Find the horizontal distance travelled when the pole vaulter lands. (3 marks)

b Given that the pole vaulter is at his greatest height halfway between leaving the ground and landing, find the greatest height of the pole vaulter. (3 marks)

For a jump to be successful, the pole vaulter must clear a bar of height 4.9 m.

c Calculate the range of horizontal distances from the bar that the pole vaulter can leave the ground and have a successful jump. (7 marks)

d State the effect in this model of

i modelling the pole vaulter as a particle, (1 mark)

ii making air ressistance negligible. (1 mark)

The height of a pole vaulter above the ground can be modelled using the equation h=1/60(125x - 12x^2) , where h metres is the vertical height of the pole vaulter and x metres is the horizontal distance travelled after his feet leave the ground.

a Find the horizontal distance travelled when the pole vaulter lands. (3 marks)

b Given that the pole vaulter is at his greatest height halfway between leaving the ground and landing, find the greatest height of the pole vaulter. (3 marks)

For a jump to be successful, the pole vaulter must clear a bar of height 4.9 m.

c Calculate the range of horizontal distances from the bar that the pole vaulter can leave the ground and have a successful jump. (7 marks)

d State the effect in this model of

i modelling the pole vaulter as a particle, (1 mark)

ii making air ressistance negligible. (1 mark)

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#2

(Original post by

If anyone could help it would be greatly appreciated!

**TSRUser1231**)If anyone could help it would be greatly appreciated!

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(Original post by

There are lots of parts to this, and you haven't said which pieces you are having problems with. On the basis that it may be all of it, let's start with (a). What is the value of h when the pole vaulter lands? How can this help you to find the horizontal distance travelled?

**Pangol**)There are lots of parts to this, and you haven't said which pieces you are having problems with. On the basis that it may be all of it, let's start with (a). What is the value of h when the pole vaulter lands? How can this help you to find the horizontal distance travelled?

25/12=1/5x

x=125/12

125/12 metres

but i need help with part b and part c

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#4

(Original post by

but i need help with part b and part c

**TSRUser1231**)but i need help with part b and part c

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(Original post by

Part (b) - they make it very clear in the question that the maximum height occurs halfway between leaving the ground and landing. Using your answer to (a), what is the value of x at this point? How can that get you the maximum height?

**Pangol**)Part (b) - they make it very clear in the question that the maximum height occurs halfway between leaving the ground and landing. Using your answer to (a), what is the value of x at this point? How can that get you the maximum height?

1/60(125(125/24)-12(125/24)^2)

3125/576 is this correct?

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for part c i got - h = 4.9

4.9=1/60(125x-125x^2)

294=-12x^2+125x

-12x^2+125x-294=0

a = -12 b=125 c=-294

use quadratic formula to get range 3.588m to 6.829m

do you think this is correct?

4.9=1/60(125x-125x^2)

294=-12x^2+125x

-12x^2+125x-294=0

a = -12 b=125 c=-294

use quadratic formula to get range 3.588m to 6.829m

do you think this is correct?

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#8

(Original post by

for part c i got - h = 4.9

4.9=1/60(125x-125x^2)

294=-12x^2+125x

-12x^2+125x-294=0

a = -12 b=125 c=-294

use quadratic formula to get range 3.588m to 6.829m

do you think this is correct?

**TSRUser1231**)for part c i got - h = 4.9

4.9=1/60(125x-125x^2)

294=-12x^2+125x

-12x^2+125x-294=0

a = -12 b=125 c=-294

use quadratic formula to get range 3.588m to 6.829m

do you think this is correct?

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(Original post by

I think that everything you have done is correct, and that you would have the majority of the marks for it. I think there is one last thing you need to do to give an answer in context. What are they actually asking you for?

**Pangol**)I think that everything you have done is correct, and that you would have the majority of the marks for it. I think there is one last thing you need to do to give an answer in context. What are they actually asking you for?

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#10

How did you get 3125/576?If you sub the numbers in the calculator that’s not what you get!

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#11

(Original post by

for part a i got 25/12x -1/5x^2 = 0

25/12=1/5x

x=125/12

125/12 metres

but i need help with part b and part c

**TSRUser1231**)for part a i got 25/12x -1/5x^2 = 0

25/12=1/5x

x=125/12

125/12 metres

but i need help with part b and part c

(Original post by

for part b i got 125/12/2 = 125/24

1/60(125(125/24)-12(125/24)^2)

3125/576 is this correct?

**TSRUser1231**)for part b i got 125/12/2 = 125/24

1/60(125(125/24)-12(125/24)^2)

3125/576 is this correct?

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#12

(Original post by

How did you get 3125/576?If you sub the numbers in the calculator that’s not what you get!

**JoeC1999**)How did you get 3125/576?If you sub the numbers in the calculator that’s not what you get!

(Wow, nine month old thread!)

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#13

(Original post by

That's what I get when I put it into a calculator...

(Wow, nine month old thread!)

**Pangol**)That's what I get when I put it into a calculator...

(Wow, nine month old thread!)

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