Question on height of a jump in terms of Power (from WPE chapter, JEE level)

[annjee212]

Homework Statement

How do we find the height at which a man can jump in terms of P,s, m and g; given the maximum power as 'P' and the distance of centre of mass of man from ground as 's'

Relevant Equations

P = F x v or P = W/t.

I used the formula P = 1/2 mv^2/t. Then multiplied by s on both sides to get P.s = 1/2 x mv^2/t x s
==> P.s = 1/2mv^3 and then expressed v = root(2gh) and equated.
BUT I Got h = (1/2g) x (2sP/m)^2/3 but the answer is h = (1/2g) x (4sP/m)^2/3
Any idea where I am going wrong?

 

[TSny]

Does ##\large \frac s t## give you the velocity of the man just at the instant he leaves the ground or does it give you the average velocity of the center of mass of the man during the time interval ##t##?

 

[annjee212]

Does ##\large \frac s t## give you the velocity of the man just at the instant he leaves the ground or does it give you the average velocity of the center of mass of the man during the time interval ##t##?

Ohh well thank you, it gives the latter part. But then wouldnt it be (s/t)/2 right? And then we rightly get 4 in the numerator. Yep, thanks!

 

[TSny]

Ohh well thank you, it gives the latter part. But then wouldnt it be (s/t)/2 right? And then we rightly get 4 in the numerator. Yep, thanks!

Yes, s/t is the average velocity. If you can assume the acceleration of the center of mass is constant during the jump, then the average velocity during the jump is equal to half the velocity at the instant of leaving the ground. I think that's what you meant.

But I'm not sure the answer that was given is correct. If P = W/t, where W is the work done by the man in jumping, then this work does not get entirely transformed into kinetic energy. Some of the work goes into lifting the center of mass of the man while jumping. The answer given doesn't take this into account.

 

[TSny]

Something else that's confusing is that the problem statement says s is the "distance of the center of mass of the man from the ground". I think they probably meant to say that s is the vertical displacement of the center of mass of the man during the jump (until his feet leave the ground).

 

[haruspex]

I used the formula P = 1/2 mv^2/t.

What is v here? What is the man's velocity at the start? What is the velocity at maximum height?

You quote P=Fv. We can take P as constant, since it is the limiting constraint. I would expand F and solve to find height as a function of time, but there might be an easier way.