(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

After i get the formula x = Vo^{2}sin(2θ)/g, I was told that I can take the derivative of x and let that equal to zero to get the max range of the projectile. Why? What does taking the derivative do in order to help us find the max angle? I know that the value of the derivative at the maximum height of the traj. would be 0, but why is that significant?

2. Relevant equations

x = Vo^{2}sin(2θ)/g => dx/dθ = Vo^{2}/g * 2cos2θ

**Physics Forums - The Fusion of Science and Community**

# Projectile motion - max range

Have something to add?

- Similar discussions for: Projectile motion - max range

Loading...

**Physics Forums - The Fusion of Science and Community**