# Thread: Range of a projectile

1. ## Range of a projectile

Hey guys, I tried to post this on Physicshelpforum, but LaTex is dead.

Should be a simple solution....but....it's eluding me.

A batter hits a ball with an initial velocity of $v_0$ of $100 \text{ f/s }$ at an angle $\theta$ to the horizontal.

An outfielder catches the ball $200 \text{ feet }$ from home plate.

Find $\theta$ if the range of a projectile is given by the formula:

$R=\frac{1}{32}v_0^2 \sin 2 \theta$

Here's what I've done so far.

$200=\frac{1}{32}(100)^2 \sin 2 \theta$

$6400=10000 \sin 2 \theta$

$\frac{16}{25}=\sin 2\theta$

$\frac{16}{25}=2\sin \theta \cos \theta$

$\frac{8}{25}=\sin \theta \cos \theta$

Now, I can't seem to boil this thing down to just one trig function. I know the answer. $\theta \approx 20^{\circ}$. I just need to know how to get to it.

Any ideas?

2. Originally Posted by masters
Hey guys, I tried to post this on Physicshelpforum, but LaTex is dead.
Blasphemy!

Originally Posted by masters
Should be a simple solution....but....it's eluding me.

A batter hits a ball with an initial velocity of $v_0$ of $100 \text{ f/s }$ at an angle $\theta$ to the horizontal.

An outfielder catches the ball $200 \text{ feet }$ from home plate.

Find $\theta$ if the range of a projectile is given by the formula:

$R=\frac{1}{32}v_0^2 \sin 2 \theta$

Here's what I've done so far.

$200=\frac{1}{32}(100)^2 \sin 2 \theta$

$6400=10000 \sin 2 \theta$

$\frac{16}{25}=\sin 2\theta$
Let's start from here. Take the inverse sine of both sides:
$\displaystyle sin^{-1} \left ( \frac{16}{25} \right )$

$\displaystyle \theta = \frac{1}{2} sin^{-1} \left ( \frac{16}{25} \right ) = 2 \theta$

Taking the inverse sine of 16/25 only gives you a reference angle. So
$\displaystyle \frac{1}{2}(180 - (\text{reference angle})) \approx 70^o$

is also a solution.

-Dan

3. Well, Dan. It makes perfect sense to me now.

I was too eager to convert that $\sin 2\theta$, thus muddying the waters a bit. Thanks for enlightening me.