Projectile motion football question

[mauritzhansen]

Hi folks,

Need some help on a projectile motion question.

Homework Statement

Given:
- a football kicker can give the ball an initial speed of 25m/s
- kicker wants to score a field goal 50m in front of goalposts
- horizontal bar of goalposts is 3.44m above the ground

Question:
What are the a.) least and b.) greatest elevation angles at which he can kick the ball to score a field goal?

Homework Equations

none

The Attempt at a Solution

I have done the following:
1.) y = y₀ + (v₀sinθ)t - 0.5(9.81)t²
so, 3.44 = (25sinθ)t - 4.91t²
2.) x = x₀ + (v₀cosθ)t
so, 50 = (25cosθ)t
so, t = 2/cosθ

I then substituted eq. 2 into eq. 1, and got the following:
3.44 = 50tanθ - 19.62sec²θ

but I am stuck here.

Could someone please tell me if I am on the right track? Is there a way to solve the above equation, or am I approaching this in the wrong way?

Thanks.

 

[Chrisas]

Don't have time to check it now. But here is what I would try. There is a trig identity that will turn sec^2 into tan^2. You would end up with a quadratic in tan that could be solved for two roots. That would get you the two angles. In other words, change sec^2 to tan^2 using the identity and them make a substitution like z = tan(theta). Solve for the two values of z and then take the inverse tangent.

 

[mauritzhansen]

Thanks Chrisas! I did look at my trig identities, but for some reason I did not see this.

I got my two angles (31 degrees and 63 degrees), which I know are correct.

 

[kuruman]

There is another way that might be easier. Let x = cos\theta. Express the tangent as \frac{\sqrt{1-x^{2}}}{x} and you know what to do with the secant. Isolate the radical on one side of the equation and square both sides to get rid of it. You should end up with a fourth degree equation, but one that you can solve because the odd powers in x are missing. This means that you can use the quadratic formula to solve for x². Throw out any negative roots for x² as unphysical. Get the angle from the cosine.

 

[mauritzhansen]

Hi kuruman - thanks for this alternative. I had been trying for a while to get a single trig function, but failed. In hindsight I cannot believe I did not see it. More practice required. ;-)

 

[kuruman]

Hi kuruman - thanks for this alternative. I had been trying for a while to get a single trig function, but failed. In hindsight I cannot believe I did not see it. More practice required. ;-)

Think of it this way: What I call the auxiliary trig functions (tan, cot, sec, csc) can all be cast in terms of sin and cos and then sin can be converted to cos if needed. From my experience, this general method will see you through most of the time. If not, the next best thing is to try the half-angle trig identities.

 

[mauritzhansen]

That is helpful - I will certainly keep this in mind. Thanks.