# Power output by basketball player when jumping

1. Nov 18, 2014

### Jazz

1. The problem statement, all variables and given/known data

A $\small{105\!-\!kg}$ basketball player crouches down $\small{0.400\ m}$ while waiting to jump. After exerting a force on the floor through this $\small{0.400\ m}$, his feet leave the floor and his center of gravity rises $\small{0.950\ m}$ above its normal standing erect position.

What was his power output during the acceleration phase?

$m = 105\ kg$

$d = 0.400\ m$

$h = 0.950\ m$

2. Relevant equations

$P = \frac{W}{t}$

$\Delta PE_g = \Delta KE$

3. The attempt at a solution

The problem disregards the work done by the player when first crouches, so the work I tried to find is just when he accelerates upwards (jumps).

First we know that: $P_{output} = \frac{W_{output}}{t} = \frac{F_{exerted}d}{t}$

The velocity with which he left the floor is the same to that just before he touches the floor when landing; and after having fallen a height $h$:

$\Delta PE_g = \Delta KE$

$mgh = \frac{1}{2}mv^2$

$v= \sqrt{2gh} = \sqrt{(2)(9.8\ m/s^2)(0.950\ m)} = 4.32\ m/s$

Assuming that the velocity was obtained after a net constant acceleration along a distance $d$, then:

$a = \frac{v^2}{2d} = \frac{2gh}{2d} = \frac{(9.8\ m/s^2)(0.950\ m)}{0.400\ m} = 23.3\ m/s^2$

Then, I try to find the average force exerted:

$F_{exerted} = F_{net} + w$

$F_{exerted} = m(a + g)$

$F_{exerted} = (105\ kg)[23.3\ m/s^2 + 9.8\ m/s^2] = 3473\ N$

Then, I tried to find the time:

$t = \frac{v-v_0}{a} = \frac{4.32\ m/s}{23.3\ m/s^2} = 0.185\ s$

And finally the power:

$P_{output} = \frac{F_{exerted}d}{t} = \frac{(3473\ N)(0.400\ m)}{0.185\ s} = 7.49\ kW$

But the answer provided by the textbook says that the power must be $8.93\ kW$. After playing around with the numbers, I noticed that what makes the difference between both results is the way the time was calculated. According to the textbook, the time is:

$t= \sqrt{\frac{2d}{(a + w)}} = \sqrt{\frac{2(0.400\ m)}{(23.3\ m/s^2 + 9.8\ m/s^2)}} = 0.156\ s$

My question is: why should $g$ be added to the $a$? It's like considering an absolute acceleration exerted by the player, but that should underestimate the time the force was exerted.

In a hypothetical situation, if I were pushing a box with an acceleration of $1\ m/s$ through a distance $d$, and it happens that I know friction is exerting an acceleration of $3\ m/s$ in the opposite direction, the time I spent pushing the box is dependent of $a$ but not $a + a_{friction}$. The latter will yield a shorter time than that I really spent pushing the box (and hence, a greater power output).

I'm puzzled.

2. Nov 19, 2014

### Staff: Mentor

Hi Jazz. You have presented your case well. I agree with you that the textbook solution is wrong.

Last edited: Nov 19, 2014
3. Nov 19, 2014

### haruspex

The question does not make it clear that the force is constant, while the power to be determined is an average. If that is correct I agree with your answer.
More realistically for muscle, power is constant. But that leads to quite a messy equation, with no solution in closed form.

4. Nov 19, 2014

### Jazz

The problems involving force that I've been dealing with so far just consider average acceleration. I know that this does not happen in real life (at least not in situations as in the problem) but I get very confused trying to understand how a problem with changing acceleration must be treated (as far as I understand, with help of calculus I will be able to do so).

In this case, if I were asked to find a constant power output, does it mean that the magnitude of the force gets reduced, but the time over which it acts gets longer along the distance $d$, 'fitting' an equation like this $\frac{\Delta F}{\Delta t}d$?

If there is no closed form for calculating constant power, is it calculated by taking shorter distances $d$ along which a force $F$ acts during a time $t$ and then add them, or by taking shorter times over which it acts and add them?

If I were asked to find

Last edited: Nov 19, 2014
5. Nov 19, 2014

### haruspex

Not sure you've understood what I mean by 'no closed form'. Using calculus, you get an equation for P which involves P in 'incompatible' ways. Specifically, I got $mg(d+h) = -P\sqrt{\frac{2h}{g}} - \frac{P^2}{mg^2}\ln\left(1-\frac {mg}{P}\sqrt{2gh}\right)$ , but that's so horrible I suspect I made an error. At least it's dimensionally correct.

6. Nov 19, 2014

### Jazz

Roughly speaking, I understood a closed form as a nice formula used to solve 'something' that seems to be messy. Maybe it's too rough.

I hope to get into the math soon to understand how to get the equation you got. Just for curiosity, $mg(d+h)$ is considering the distance he crouches, why is that?

So when considering constant P, both I and the textbook are wrong, right?

7. Nov 19, 2014

### haruspex

That's the total work done, no?
Without completely solving it, I can't be certain it won't give the textbook answer, but it seems unlikely. As I wrote, my equation could be wrong.

8. Nov 19, 2014

### Jazz

Yep, I was just overthinking about it.

(:

Thanks for helping me to see the light with this.