# Homework Help: Football related Projectile problem!

1. Jan 25, 2008

### Go Angels

1. The problem statement, all variables and given/known data
A placekicker must kick a football from a point 36.0 m (about 40 yards) from the goal, and half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 48.0° to the horizontal.
(a) By how much does the ball clear or fall short of clearing or fall short of clearing the crossbar? (Enter a negative answer if it falls short.)

2. Relevant equations
Possibly:
vxi = vi cos theta
vyi = vi sin theta
change(y) = viy(t) + .5(a)(t)^2

3. The attempt at a solution
20 sin 48(t) - 4.9(t)^2
= 26.75-4.9(1.8)^2
change(y) = 26.75 - 15.88 =10.87

10.87-3.05 = 7.82 m. Which is the wrong answer.

2. Jan 25, 2008

### Staff: Mentor

Start by figuring out the time it takes for the ball to reach the goal. Hint: Consider the horizontal motion.

3. Jan 25, 2008

### Go Angels

time = (vi sin theta) / g

Is that the correct equation?

So it's (20 sin 48) / 9.8 = 1.52

I plugged that in and obtained 8.21 m as my finals answer, which is incorrect. Can some please tell me what I did wrong?

4. Jan 25, 2008

### Staff: Mentor

No, that's not correct. Hint: The horizontal component of motion is unaccelerated. (Gravity only acts downward.)

What's the speed in the horizontal direction?

5. Jan 25, 2008

### Go Angels

(20 cos 48) / 9.8?

6. Jan 25, 2008

### Staff: Mentor

Why divide by 9.8? (Check your units.)

7. Jan 25, 2008

### Go Angels

Opps, I thought I was still looking for time.

Why would I need to know the speed though, since the velocity is already given.

I basically just plug everything into the R = (vi^2 sin 2theta)/g equation right?

8. Jan 25, 2008

### Staff: Mentor

That's what we're heading towards calculating.

Then when I asked for the speed in the horizontal direction you should have answered without hesitation.

Nope. That's a range equation.