Question

When making a throw to first base, a third baseman is typically about 130 feet from first base. with the equation
d = (v^2/16) (sin(theta)cos(theta))
what is the smallest angle a ball can be thrown with an initial velocity of 120 ft/second

Answer 1

You'll need the identity \(\sin(2x) = 2\sin(x)\cos(x)\) with \(x = \theta\).

Answer 2

Because the angle \(\theta = 0^\circ\) would give \(d = 0\), and \(d\) as a function of \(\theta\) is continuous, you know that the lowest \(\theta\) for which \(d = 130\) is the lowest \(\theta\) for which\(d \geq 130\).

Answer 3

So you're looking to solve \(130 = \frac{v^2}{16}\sin(\theta)\cos(\theta) = \text{...}\) (use the identity).

Answer 4

That's as far as I've been getting. Identities stump me.

Answer 5

You have \(\sin(2\theta) = 2\sin(\theta)\cos(\theta); \sin(\theta)\cos(\theta) = \frac12\sin(2\theta)\) which you are to substitute into the equation.

Answer 6

You should get
\[130 = \frac{v^2}{16}\frac{1}{2}\sin(2\theta) = \frac{v^2}{32}\sin(2\theta) \text{ [divide by } \frac{v^2}{32}\text{]}\\
\sin(2\theta) = 130\frac{32}{v^2} \text{ [take arcsin on both sides]}\\
2\theta = \arcsin(130\frac{32}{v^2})\\
\theta = \frac12\arcsin(130\frac{32}{v^2}) = \frac12\arcsin(130\frac{32}{120^2}) \approx 8.4^\circ
\]

Answer 7

\[.144 = \frac{ 1 }{ 2 }\sin(2\Theta)\]
\[.288 = \sin(2\Theta)\]
\[16.738 = 2\Theta \]
\[8.369\]degrees = theta

Answer 8

Dangit! I had it figured out just couldn't put it in as fast as you haha.

Answer 9

@JoelSjogren Thank you again for all the help!

Answer 10

Good job and no problem!