# Homework Help: Difficult Projectile problem

1. Oct 11, 2006

### fro

The question is:
"A golf ball travels 418.78 meters. Assuming no air resistance and that the ball was shot at an optimum angle, what was the initial horizontal component of the ball's velocity?"

I am assuming 45 degrees to be the optimum angle. I am unsure if 418.78m is the length of the adjacent side (horizontal distance) or the length of the hypotenuse?

If I take 418.78m as the length of the hypotenuse, then the horizontal distance should be 296.122m (418.78 X cosine 45).

Since I do not know the time or the initial velocity, how can I solve this problem?

2. Oct 11, 2006

### neutrino

418.78 is the horizontal range. Now, since you know the range and the projection angle, you can caluculate the initial velocity using the formula for range.

3. Oct 11, 2006

You do not have to know the time. The total distance that the ball travelled is enough. Use the equation for displacement in the x-direction, which is $$x(t)=v_{0x}\cos(45)\cdot t$$. As stated, you know the distance, so plug it into the equation. The next equation you need to use is the equation of vertical displacement, i.e. in the y-direction, which is $$y(t)=v_{0y}\sin(45)\cdot t-\frac{1}{2}\cdot g \cdot t^2$$. Further on, you know what y equals when the ball reaches its final distance, i.e. when it falls onto the ground. I'm sure you know how to proceed now.(Hint: eliminate the time t from the equation.)

4. Oct 11, 2006

### fro

So,
418.78m = Vx X time ?

Vx = initial horizontal velocity

Since horizontal acceleration is 0, 1/2*a*t^2 is also 0 and can be left out.

5. Oct 11, 2006

### neutrino

If you've not come across the formula I mentioned before, I guess it's better to use the method suggested by radou. :)

6. Oct 11, 2006

### fro

418.78 = Vix * cos(45) * t
418.78 = Vix * 0.707 * t
t = 592.24/Vix.

Can I plug the above value of t in the y-displacement equation?

If so then,
0 = Viy * sin(45) * (592.24/Vix) * - [0.5 * g * (592.24/Vix)^2]
0 = Viy * 0.707 * (592.24/Vix) - (1753741.088/Vix^2)

Can I cancel out the Viy and Vix since I know that the vertical and horizontal components must be equal (because it is 45 degress)?

7. Oct 11, 2006

### neutrino

It's not Vix cos(45) , Vix = Vi cos(45), where Vi is the initial velocity. But you're right, Vix = Viy, in this case.

8. Oct 11, 2006