The range R of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at t = 0 with initial speed vi at an angle θ above the horizontal. The range is

R = vi2sin2θ/g. What's the launch angle θ at which the maximum range occurs?

When is sinZ maximum? Z=90 degrees. So what is theta at max range?

To find the launch angle θ at which the maximum range occurs, we need to find the angle that maximizes the range equation R = vi^2 * sin(2θ)/g.

To do this, we can take the derivative of the range equation with respect to θ and set it equal to zero, and then solve for θ.

Let's start by using the chain rule to take the derivative. The derivative of the range equation with respect to θ is:

dR/dθ = (vi^2 * cos(2θ))/g

Setting this derivative equal to zero, we have:

(vi^2 * cos(2θ))/g = 0

We can now solve for θ:

cos(2θ) = 0

Using the identity cos(2θ) = 2cos^2(θ) - 1, we have:

2cos^2(θ) - 1 = 0

Rearranging this equation, we get:

cos^2(θ) = 1/2

Taking the square root of both sides, we have:

cos(θ) = ±√(1/2)

Solving for θ, we find:

θ = ±45°

Therefore, the launch angle at which the maximum range occurs is either 45 degrees or 135 degrees.