# Homework Help: Why does this system of equations lead to inaccurate results?

1. Oct 4, 2016

### k_squared

1. The problem statement, all variables and given/known data
A soccer ball is kicked. If there is a 3 meter high fence, that is 6 meters away, compute the angle of the kick and the magnitude of the velocity of a ball.

2. Relevant equations
a_y=-g
v_y=-gt+v_y_0
s_y=-(1/2)gt^2+v_y_0(t)

v_x is constant: we neglect air resistance.
6=v_x_0t

3. The attempt at a solution

Because the maximum height is 3, the direction of v_y changes from positive to negative at this point.
0= -gt+v_y_0
3=-.5gt^2+t*v_y_0

We solve this system of equations to yield the initial speed in the y direction, and the time at which the ball passes over the fence. We then use the equations for the position on the x axis: 6=tv_0_x and compute V_0_x.

We use arc-tangent to get the angle from the two speeds. The angle is wrong. My maximum speed is close, but still wrong.

What is the problem with this set of equations such that it will not yield a correct answer?

2. Oct 4, 2016

Your equations look correct. Show a little more work please such as the answers that you got and maybe we can see if you got the correct answer or not.

3. Oct 4, 2016

### Staff: Mentor

Your problem statement sets the scene, but it doesn't state the goal. Is the ball meant to clear the fence? If so, it doesn't state that it must do so at the top of the ball's trajectory: the ball could still be rising at that point, or descending. In fact, as given there are infinitely many solutions (given a suitably ideal ball ).

4. Oct 4, 2016

5. Oct 4, 2016

### haruspex

I assume the question wants you to find the angle for the least initial speed.
As gneill points out, that's just an assumption. The question as you have written it does not state that. Try dropping it.
(The official answer is greater for the angle and less for the speed than you obtained, right?)

6. Oct 4, 2016

### k_squared

I'm sorry. The question states: *FIND THE MINIMUM INITIAL VELOCITY" for the ball to clear the fence. I should have made this clearer!

7. Oct 5, 2016

### ehild

Supposing that the maximum height is 3 m and the distance of the fence is 6 m, what did you get for the tangent of the angle?

8. Oct 5, 2016

### k_squared

45 degrees. The V_y_0 and V_x_0 are very very close.

9. Oct 5, 2016

### ehild

Then drop the condition that 3 m is the maximum height. The ball must clear the point (6,3). Eliminate the time and use the equation of the trajectory of the ball. Write v02 in terms of θ, and find its minimum.

Last edited: Oct 5, 2016
10. Oct 5, 2016

### k_squared

But if it were the slowest possible launch speed, wouldn't that mean that the top of the fence is at the vertex?

11. Oct 5, 2016

### ehild

Can you prove it? It is not sure.

You certainly know the function y(x), the height of the projectile at distance x (https://en.wikipedia.org/wiki/Trajectory_of_a_projectile)

$y=x\tan(\theta)-\frac{gx^2}{2(v_0 \cos (\theta))^2}$

In our case y=3 m if x=6 m. This gives you a relation between the initial speed v0 and the angle θ. At what angle is v02 minimum?

12. Oct 5, 2016

### haruspex

If you launch it at a slightly steeper angle than that you will need a bit more vertical speed to get it over. But it will take longer to reach the fence, so you do not need as much horizontal speed. Which wins? Do the math and find out.