Momentum of a Punt

1. Feb 7, 2017

kq6up

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

A punter drops a ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18m/s at an angle of 55$^{\circ}$ above the horizontal. What is the impulse delivered by the foot (magnitude and direction)?

2. Relevant equations
$p_b+p_f=p^{\prime}_{b}+p^{\prime}_{f}$
$v^2=v_{o}^2+2gh$

3. The attempt at a solution
$\sqrt{2gh}=4.43m/s$

Would not one need to know the mass of the football to solve this problem? I find it uncomfortable and strange that the mass of the football is not included in this problem. Am I correct in this assertion?

Thanks,
KQ6UP

2. Feb 7, 2017

TSny

Yes, you need the mass in order to get a numerical value for the magnitude of the impulse. You might have to settle for specifying the answer in terms of the mass.

3. Feb 7, 2017

jbriggs444

You can Google for the mass of a regulation football. (0.40 to 0.43 kg).

4. Feb 8, 2017

kq6up

Ok, I used .43kg, and I get:
$v=\sqrt{2gh}$ implies that $v=4.43m/s$.

I used $I=\Delta \boldsymbol{P}=\boldsymbol{P}-\boldsymbol{P}_o$

Then componentized:

$\boldsymbol{P}_o=-1.90kg\cdot m/s \cdot \hat{y}$

$\boldsymbol{P}=18 m/s \cdot .43 kg \cdot (\cos 55^o \hat{y}+\sin 55^o \hat{x})$

$\therefore$ $\Delta \boldsymbol{P}$ is a vector pointing 61.7$^o$ with a magnitude of 9.36$kg \cdot m/s$. Is this correct?

Thanks,
KQ6UP

Last edited: Feb 8, 2017
5. Feb 8, 2017

TSny

Your answer looks correct. There appears to be a typographical error where your unit vectors $\hat{x}$ and $\hat{y}$ need to switch places in the expression for P.

6. Feb 9, 2017

kq6up

Yes, that is a typo. Thanks,

KQ6UP

7. Feb 9, 2017

kq6up

Here it is fixed for the record.