- #1
leitz
- 8
- 0
Homework Statement
After i get the formula x = Vo2sin(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?
Homework Equations
x = Vo2sin(2θ)/g => dx/dθ = Vo2/g * 2cos2θ