# Projectile motion with people moving

1. Oct 29, 2006

### sunbunny

I'm having trouble with this problem as well:

Quarterback Fred is going to throw a pass to tight end Doug. Doug is 20 m in front of Fred and running straight away at 6.0 m/s when Fred throws the 500 g football at a 40 angle. Doug catches the ball without having to alter his speed and runs for the game-winning touchdown.

How fast did Fred throw the ball?

I don't even know where to start with so so if anyone can help me get started that would be great.

Thanks

2. Oct 29, 2006

### geoffjb

Look at these two equations and figure out how and why they work:

$$6t+20=v_{0}\cos(\theta)t$$

$$v_{0}\sin(\theta)t=\frac{gt^{2}}{2}$$

Solve each in terms of $v_{0}$, equate one equation to the other, solve for t, and then plug it into a formula to find $v_{0}$.

3. Oct 29, 2006

### sunbunny

Okay, so when i solved for vo i for:

vo= 6t+20/cost and vo= gt^2/2sint

then i made these equations equal to each other:

6t + 20/cost = gt^2/sint

but when i solve for t, the value that i get donesn't work. if you could explain how to solve for t properly that would be geat.

4. Oct 29, 2006

### geoffjb

The equation you end up with is a quadratic. Manipulate it to form $$ax^{2}+bx+c=0$$, and solve for x. You will get two roots$-$only one will be applicable.

Last edited: Oct 29, 2006
5. Oct 29, 2006

### sunbunny

okay this time when i did that i got:

12sint^2 + 40sint=gcost^3

-gcost^3+ 12sint^2 + 40sint=0

t(-gcost^2 + 12sint =40sin)=0

i solved for inside the brackets using the quadratic formula and i got 2.43 for my time and then i put it back into one of the equations and i got 78m/s however i still got the wrong answer.

I was wondering if you could please check over my work to make sure that i'm not missing anything. Thanks!

6. Oct 29, 2006

### geoffjb

2.43 s is the correct time. Which equation did you plug it back into?

7. Oct 30, 2006

### sunbunny

i put it into the 6t+20/cost equation

8. Oct 30, 2006

### geoffjb

Put it into:

$$y=v_{0}\sin(\theta)t-\frac{gt^{2}}{2}$$

Remember, at the time you've specified, y will have a certain value.

Your question is solved with this equation.

9. Oct 30, 2006

### sunbunny

okay i got it now, thank you for your help!