Impulse of Punt Homework: Mass Needed?

[kq6up]

Homework Statement

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)?[/B]

Homework Equations

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

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[/B]

 

[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.

 

[jbriggs444]

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

 

[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

 

[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.

 

[kq6up]

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.

Yes, that is a typo. Thanks,

KQ6UP

 

[kq6up]

Here it is fixed for the record.

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:

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

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

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

Thanks,
KQ6UP