# Parametric Equations Word Problem

1. Jul 20, 2010

1. The problem statement, all variables and given/known data
Consider a projectile launched at a height of h feet above the ground at an angle θ with the horizontal. If the initial velocity is v0 feet per second, the path of the projectile is modled by the parametric equations
x=(v0cos θ)t and y=h + (v0 sin θ)t-16t2.

The center-field fence in a ballpark is 10 feet high and 400 feet from home plate. The baseball is hit 4 feet above the ground. It leaves the bat at an angle of θ degrees with the horizontal at a speed of 100 miles per hour.

Find the minimum angle required for the hit to be a home run

2. Relevant equations

3. The attempt at a solution

$$x=(146.67\cos\theta)t$$
$$y=3+(146.67\sin\theta)t-16t^2$$

by the question when x=400, y>10 the ball will pass over the fence

so if I solve for the angle θ when x=400 and y=10 the angle I get should be the minimum passable.

$$400=(146.67\cos\theta)t$$
$$10=3+(146.67\sin\theta)t-16t^2$$

$$\frac{400}{146.67\cos\theta}=t$$

sub that into the other equation

$$y=3+(146.67\sin\theta)(\frac{400}{146.67\cos\theta})-16(\frac{400}{146.67\cos\theta})^2$$

I can get a common denominator and get it to

$$7=400(146.67)^2\cos^2\theta\sin\theta-16(400)^2\cos\theta$$

but I'm not quite sure what to do after here to solve for θ.

Last edited: Jul 20, 2010
2. Jul 20, 2010

### eumyang

Is it 3 or 4 in the 2nd equation? Your original problem says 4 feet.

Assuming that it's 3:
(I assume you meant to put a 10 in for y.) From here, try writing in terms of tan θ. I see a tan θ and a sec2 θ "hidden" in this equation, and you can use the pythagorean identity 1 + tan2 θ = sec2 θ. That way you'll end up with a quadratic in tan θ.

69

3. Jul 20, 2010

Yeah, its suppose to be 3. sorry.

so I'd have

$$10=3+(146.67\sin\theta)(\frac{400}{146.67\cos\theta})-16(\frac{400}{146.67\cos\theta})^2$$

$$7=(\frac{(400\sin\theta)}{\cos\theta})-(\frac{16(400)^2}{146.67^2\cos^2\theta})$$

$$7=400\tan\theta-\frac{16(400)^2\sec^2\theta}{146.67^2}$$

Then...

$$7=400(146.67)^2\tan\theta-16(400)^2\sec\theta$$

$$0=-16(400)^2\tan^2\theta+400(146.67)^2\tan\theta-16(400)^2-7$$

Expand, factor out negative

$$0=-(2560000\tan^2\theta-8604835.56\tan\theta+2559993)$$

$$\tan\theta=\frac{8604835.56\pm\sqrt{-8604835.56^2-4(2560000)(2559993)}}{2(2560000)}$$

For the minus I get .3299 for the plus I get 3.031 the plus has way to high of an angle when I run an arctan on it and .3299 comes out as 18.27 degrees but this is wrong because I can check by plugging it back into the equation. and when x=400, y does not equal 10. It's close though. θ is suppose to equal "about 19.4 degrees" by the answer key.

4. Jul 20, 2010

### eumyang

You forgot to multiply the left side by 146.672 here. As it is, I wouldn't multiply both sides by 146.672 at all. I would simplify the coefficient of sec2 θ first:

$$7 &= 400 \tan \theta - \frac{16(400)^2 \sec^2 \theta}{146.67^2}$$

$$7 &= 400 \tan \theta - \frac{14400}{121} \sec^2 \theta$$

(440/3 ≈ 146.67.)
At this point, if you want, you could multiply both sides by 121 and then use the pythagorean identity.

69

5. Jul 20, 2010