# Range of a projectile

• Mar 15th 2011, 10:13 AM
masters
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 $\displaystyle v_0$ of $\displaystyle 100 \text{ f/s }$ at an angle $\displaystyle \theta$ to the horizontal.

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

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

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

Here's what I've done so far.

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

$\displaystyle 6400=10000 \sin 2 \theta$

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

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

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

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

Any ideas?
• Mar 15th 2011, 10:56 AM
topsquark
Quote:

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

Blasphemy!

Quote:

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

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

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

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

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

Here's what I've done so far.

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

$\displaystyle 6400=10000 \sin 2 \theta$

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

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

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

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

is also a solution.

-Dan
• Mar 15th 2011, 11:21 AM
masters
Well, Dan. It makes perfect sense to me now.

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