# Range of a Projectile

• Mar 15th 2010, 03:36 PM
touille89
Range of a Projectile
ok so I'm having some problems with this problem, it is an optimization problem and the paper says this: The range R of a projectile whose muzzle velocity in meters per second is v, and whose angle of elevation in radians is theta, is given by R=(v^2*sin(2*theta))/g, where g is the acceleration due to gravity. Which angle of elevation gives the maximum range of the projectile?
if i could get some help figuring out how to do this, it would help me out a lot, thanks
• Mar 15th 2010, 04:23 PM
skeeter
Quote:

Originally Posted by touille89
ok so I'm having some problems with this problem, it is an optimization problem and the paper says this: The range R of a projectile whose muzzle velocity in meters per second is v, and whose angle of elevation in radians is theta, is given by R=(v^2*sin(2*theta))/g, where g is the acceleration due to gravity. Which angle of elevation gives the maximum range of the projectile?
if i could get some help figuring out how to do this, it would help me out a lot, thanks

find $\displaystyle \frac{dR}{d\theta}$ , set it equal to 0 and maximize.

remember to treat v and g as constants.
• Mar 15th 2010, 05:00 PM
touille89
so I've got the derivative now and I've set it equal to zero, but the way that i went about solving for zero i don't think it worked, i checked by putting some numbers in for v and g and it didn't give me a zero, i used the double angle formula for cosine because that was the only way that i could figure out how to do it. The derivative that i have gotten is (v^2*cos(2*theta))/g
• Mar 15th 2010, 05:08 PM
skeeter
Quote:

Originally Posted by touille89
so I've got the derivative now and I've set it equal to zero, but the way that i went about solving for zero i don't think it worked, i checked by putting some numbers in for v and g and it didn't give me a zero, i used the double angle formula for cosine because that was the only way that i could figure out how to do it. The derivative that i have gotten is (v^2*cos(2*theta))/g

$\displaystyle R = \frac{v^2 \sin(2\theta)}{g}$

$\displaystyle \frac{dR}{d\theta} = \frac{2v^2\cos(2\theta)}{g}$

$\displaystyle \frac{dR}{d\theta} = 0$ when $\displaystyle \cos(2\theta) = 0$

... note that $\displaystyle \cos(90^\circ) = 0$
• Mar 15th 2010, 05:24 PM
touille89
wow, i can't believe i forgot that, thanks so much for helping me out with this, i figured it out, thanks!!!