# Thread: Maximizing The Range Of A Projectile

1. ## Maximizing The Range Of A Projectile

The range of a projectile is R = v0^2*sin(2*theta)/g where v0 is its initial velocity, g is the acceleration due to gravity and is a constant, and theta is its firing angle. Find the angle that maximizes the projectile's range.

I know the answer is 45 degrees, but I'm confused as to how to arrive at that answer using optimization.

2. Originally Posted by Fred119
The range of a projectile is R = v0^2*sin(2*theta)/g where v0 is its initial velocity, g is the acceleration due to gravity and is a constant, and theta is its firing angle. Find the angle that maximizes the projectile's range.

I know the answer is 45 degrees, but I'm confused as to how to arrive at that answer using optimization.

Derivate the function $R(\theta)=\frac{v_0^2\sin(2\theta)}{2}$ wrt $\theta$ and equate the derivative to zero:

$R_{\theta}=v_0^2\cos(2\theta)=0\,\Longrightarrow\, 2\theta=\pi \,\,radians=90\,\, degrees\,$ $\Longrightarrow\,\theta=\frac{\pi}{2}\, radians=45\,\,degrees$

Tonio

3. Originally Posted by tonio
Derivate the function $R(\theta)=\frac{v_0^2\sin(2\theta)}{2}$ wrt $\theta$ and equate the derivative to zero:

$R_{\theta}=v_0^2\cos(2\theta)=0\,\Longrightarrow\, 2\theta=\pi \,\,radians=90\,\, degrees\,$ $\Longrightarrow\,\theta=\frac{\pi}{2}\, radians=45\,\,degrees$

Tonio
Why did you rewrite the function over 2 instead of g? Thank you.

4. Originally Posted by Fred119
Why did you rewrite the function over 2 instead of g? Thank you.

My bad, as simple as that...but the derivative vanishes in the very same point as before, so nothing REALLY important changes in the proof.

Tonio