Parametric Equations Word Problem

[themadhatter1]

Homework Statement

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

Homework Equations

The Attempt at a Solution

So your basic equations are

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.

 

[eumyang]

So your basic equations are

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

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

Assuming that it's 3:

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

(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

 

[themadhatter1]

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

Yeah, its suppose to be 3. sorry.

so I'd have

<br /> 10=3+(146.67\sin\theta)(\frac{400}{146.67\cos\theta})-16(\frac{400}{146.67\cos\theta})^2<br />

7=(\frac{(400\sin\theta)}{\cos\theta})-(\frac{16(400)^2}{146.67^2\cos^2\theta})<br />

<br /> 7=400\tan\theta-\frac{16(400)^2\sec^2\theta}{146.67^2}<br />

Then...

<br /> 7=400(146.67)^2\tan\theta-16(400)^2\sec\theta<br />

<br /> 0=-16(400)^2\tan^2\theta+400(146.67)^2\tan\theta-16(400)^2-7<br />

Expand, factor out negative

<br /> 0=-(2560000\tan^2\theta-8604835.56\tan\theta+2559993)<br />

Quadratic equation

<br /> \tan\theta=\frac{8604835.56\pm\sqrt{-8604835.56^2-4(2560000)(2559993)}}{2(2560000)}<br />

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.

 

[eumyang]

<br /> 7=400\tan\theta-\frac{16(400)^2\sec^2\theta}{146.67^2}<br />

Then...

<br /> 7=400(146.67)^2\tan\theta-16(400)^2\sec^2 \theta<br />

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

7 &amp;= 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

 

[themadhatter1]

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

7 &amp;= 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

Ok, now I see how to do it. Thanks!