Power output by basketball player when jumping

In summary, a basketball player with a mass of 105 kg crouches down 0.400 m and then jumps, reaching a height of 0.950 m. The question asks for the power output during the acceleration phase, assuming constant acceleration. After analyzing the problem and calculating the power output using different methods, it is concluded that the textbook solution is incorrect and the correct answer is 7.49 kW. The question also raises the issue of how to calculate power when it is constant, with the suggestion that a closed form equation may not exist.
  • #1
Jazz
103
5

Homework Statement



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##

Homework Equations



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

##\Delta PE_g = \Delta KE##

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.
 
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  • #2
Hi Jazz. You have presented your case well. I agree with you that the textbook solution is wrong.
 
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  • #3
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
Thank you, both, for your answers.

haruspex said:
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.

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
 
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  • #5
Jazz said:
Thank you, both, for your answers.
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
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
haruspex said:
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.

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
Jazz said:
##mg(d+h)## is considering the distance he crouches, why is that?
That's the total work done, no?
Jazz said:
So when considering constant P, both I and the textbook are wrong, right?
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
haruspex said:
That's the total work done, no?

Yep, I was just overthinking about it.

haruspex said:
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.

(:

Thanks for helping me to see the light with this.
 

Related to Power output by basketball player when jumping

What is the definition of power output in basketball?

Power output in basketball refers to the amount of energy a player is able to produce when jumping. It is a measure of the player's explosive strength and is an important factor in their performance on the court.

How is power output measured in basketball?

Power output can be measured using various methods, such as force plates, jump mats, or motion capture technology. These tools measure the player's vertical jump and the amount of force they apply to the ground, providing a measure of their power output.

What factors affect power output in basketball?

The main factors that affect power output in basketball include the player's muscle strength, muscle fiber type, technique, and body composition. Other factors such as fatigue, equipment, and surface also play a role.

What are some ways to improve power output in basketball?

To improve power output in basketball, players can focus on strength training exercises that target the lower body muscles used in jumping. Plyometric exercises, such as box jumps and depth jumps, can also help improve explosive power. Proper technique and form are also crucial in maximizing power output.

Is power output more important than height in basketball?

Both power output and height play important roles in basketball performance. While power output can greatly impact a player's ability to jump and perform explosive movements, height can also provide advantages in terms of rebounding and defending. Ultimately, a combination of power and height is ideal for optimal performance on the court.

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