Why does this kinematics problem seem impossible.HELP

AI Thread Summary
The kinematics problem involves calculating the downward force exerted by a man's feet during a jump and determining if he can clear a barrier of 0.8 m height. The calculated downward force is 1715 N, leading to an acceleration of 24.5 m/s². To assess the jump height, the necessary calculations indicate that the maximum height reached would be 0.72 m, which is insufficient to clear the 0.8 m barrier. Additionally, momentum is not conserved during the jump due to the constant influence of gravity. The discussion highlights the challenges in solving the problem, particularly in determining the time and height parameters.
charmedbeauty
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why does this kinematics problem seem impossible.HELP!

Homework Statement



If the man's mass is 70 kg, if L is 0.4 m h is 0.6 m, what is the downwards force (assumed constant) exerted by his feet during the phase A to B? State any physical law or principle you use in obtaining your answer.

vi) For a man whose parameters are given in part (v), would it be possible to jump over a barrier 0.8 m high starting the jump as shown from A to B? Explain your answer.

heres a link to the problem http://www.phys.unsw.edu.au/firstyear/pdf%20files/1131T12008.pdf

Homework Equations


The Attempt at a Solution



I am really struggling with part vi)

the answer for part v) is 1715N

now i know the highest point reached is when vf=0

but I am having a hard time figuring out the time for this.

All I know is that 1716=ma

so a = 24.5 m/s

and I need to see if this acceleration is large enough for a man of mass 70kg can reach a height h=0.8

are there to many unknowns in this problem or is it just me ??

I have tried so many different ways but I keep getting stuck. please help!

also part vii)During which (if any) phases of the jump described in part (vi) is the momentum of the man conserved? Explain your answer briefly.

would it be none; since gravity is always acting on the system.??
 
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charmedbeauty said:

Homework Statement



If the man's mass is 70 kg, if L is 0.4 m h is 0.6 m, what is the downwards force (assumed constant) exerted by his feet during the phase A to B? State any physical law or principle you use in obtaining your answer.

vi) For a man whose parameters are given in part (v), would it be possible to jump over a barrier 0.8 m high starting the jump as shown from A to B? Explain your answer.

heres a link to the problem http://www.phys.unsw.edu.au/firstyear/pdf%20files/1131T12008.pdf

Homework Equations



The Attempt at a Solution



I am really struggling with part vi)

the answer for part v) is 1715N

now i know the highest point reached is when vf=0

but I am having a hard time figuring out the time for this.

All I know is that 1716=ma

so a = 24.5 m/s

and I need to see if this acceleration is large enough for a man of mass 70kg can reach a height h=0.8

are there to many unknowns in this problem or is it just me ??

I have tried so many different ways but I keep getting stuck. please help!

also part vii)During which (if any) phases of the jump described in part (vi) is the momentum of the man conserved? Explain your answer briefly.

would it be none; since gravity is always acting on the system.??
What are your answers to parts ii & iii, including how you arrived at them?

How did you arrive at your answer to part iv ?
 
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Ditto SammyS' post.
 


SammyS said:
What are your answers to parts ii & iii, including how you arrived at them?

How did you arrive at your answer to part iv ?

ohhh now I see I can use this for vo...

by the way for vicm I did

Ui+Ki=Ui+Ki

and solved for vcm=√(2gh)

Thanks so much I never thought to look at my previous answers and had been stuck for ages. Thanks.
 


SammyS said:
What are your answers to parts ii & iii, including how you arrived at them?

How did you arrive at your answer to part iv ?

Does this seem right now...

vf=\sqrt{2gh}+24.5t
vf=0

t=\sqrt{2gh}/24.5

t=0.14

xy= \sqrt{2gh}(0.14)+\frac{1}{2}(24.5)(0.14)^{2}

xy= 0.72

Therefore no -----> 0.72<0.80

Does this seem right?
 
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The acceleration you have for part iii is wrong.

The force looks right, but that's not the net force exerted upon the CoM .
 


SammyS said:
The acceleration you have for part iii is wrong.

The force looks right, but that's not the net force exerted upon the CoM .

ok I have to subtract the acceleration due to gravity so the right acceleration should be 14.7m/s2??
 
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