Homework Help: Projectile Motion football kick

1. Oct 3, 2009

Osbourne_Cox

1. A place kicker must kick a football from a point 35.0 m (about 38 yd) from the goal. As a result of the kick, the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 22.0 m/s at an angle of 52◦ to the horizontal. The acceleration of gravity is 9.81 m/s2 .

a) To determine if the ball clears the cross-bar, what is its height with respect to the crossbar when it reaches the plane of the crossbar?

b) To determine if the ball approaches the crossbar while still rising or while falling, what is its vertical velocity at the crossbar?

2. Relevant equations

3. The attempt at a solution

No clue! I haven't done physics in three years and simply don't remember where to begin! Please help! (:

2. Oct 3, 2009

w3390

First, write out the kinematic equations for each direction, both x and y. Since there is no acceleration in the x direction, you can solve for t (the time it takes for the ball to get the plane of the crossbar). You can then plug this into the equation for the y direction and solve for the height of the ball after a time t.

3. Oct 3, 2009

Osbourne_Cox

Okay, so first I used x=Vxt to get a time of 1.6 s. Was that what I was supposed to do? Now, how do I find the height using that?

4. Oct 3, 2009

w3390

That is what you were supposed to do. To find the height use the same equation you used for the x direction except, since the y direction has acceleration, you will have the extra term -(1/2)at^2.

5. Oct 3, 2009

Osbourne_Cox

Okay. I have a question...as I was doing more questions, I noticed the equations Vxi=Vicostheta and Vyi=Visintheta.

Was I supposed to use those for Vx because i just subbed in the velocity, not the x component of the velocity.

6. Oct 3, 2009

pr0blumz

Vxi = 22.0 m/s*cos(52)
Vyi= 22.0 m/s*sin(52)

7. Oct 3, 2009

Osbourne_Cox

So that gives me a time of 2.6 s instead.

8. Oct 3, 2009

Correct

9. Oct 3, 2009

Osbourne_Cox

Now, using that same equation, just for y with the extraterm -(1/2)at^2, does that give me the velocity when it is over the bar? and is the 9.8 a negative value? If so ( or if not ) could you kindly explain why? How is height determined from this value? thank you very much!

10. Oct 3, 2009

pr0blumz

use y = v_o(t)+1/2a(t)^2. Use that to find the position of the ball after 2.6 seconds. Yes, 9.8 is negative due to the acceleration of gravity

11. Oct 3, 2009

Osbourne_Cox

Okay, that gives me a velocityat 2.6 s of 11.9 if a is positive or 78.1 if a is negative....how do I know which to use?

12. Oct 3, 2009

pr0blumz

y = 22*sin(52)(2.6)+.5(-9.8)(2.6)^2

That's how your equation should look.

13. Oct 3, 2009

Osbourne_Cox

Okay, so y=11.9

What is this y value? is it the velocity at 2.6 s?

14. Oct 3, 2009

pr0blumz

Yes, @ 2.6s.

15. Oct 3, 2009

Osbourne_Cox

So would 11.9 be the answer for part b?

16. Oct 3, 2009

pr0blumz

subtract the height of the crossbar from y @ 2.6 seconds. When you do that, that will tell you by how much the ball clears

17. Oct 3, 2009

Osbourne_Cox

Ah, I'm confused! subtract 3.05 from 11.9? what will that give me?

18. Oct 3, 2009

pr0blumz

11.9 - 3.05. = The height of the ball above the crossbar.

19. Oct 4, 2009

Osbourne_Cox

I put that in for the answer and it says I'm wrong.

20. Oct 4, 2009

pr0blumz

Hoid on, let me check everything

21. Oct 4, 2009

pr0blumz

Are all your initial values correct?

22. Oct 4, 2009

Osbourne_Cox

Yes, all of the values listed in the problem are correct.

23. Oct 4, 2009

pr0blumz

Well, I got the ball will be 9.06 meters above the crossbar with a velocity of -7.95 and the negative means its falling. I had the same exact problem for homework but my values were a little different than yours. I used 2.58s as the time in your problem.

24. Oct 4, 2009

Osbourne_Cox

Those two answers are correct. Where did you get 2.58 s from?

25. Oct 4, 2009

pr0blumz

When I divided 35.0/22.0*sin(52), I kept 3 significant digits.