Question 1209422
**1. Find the time of flight:**

* The projectile hits the ground when y = 0. 
* Set the y equation to zero: 
   vt * sin(theta) - (1/2) * g * t^2 = 0
* Solve for t:
   t(vt * sin(theta) - (1/2) * g * t) = 0
   t = 0 (initial point) or t = (2 * v * sin(theta)) / g 

**2. Find the horizontal distance (range):**

* Substitute the time of flight (t = (2 * v * sin(theta)) / g) into the x equation:
   x = v * cos(theta) * (2 * v * sin(theta)) / g
   x = (v^2 * 2 * sin(theta) * cos(theta)) / g

**3. Use the trigonometric identity:**

* Recall the double-angle formula for sine: 
   sin(2 * theta) = 2 * sin(theta) * cos(theta)

* Substitute into the range equation:
   x = (v^2 * sin(2 * theta)) / g

**4. Find the maximum distance:**

* The maximum distance occurs when sin(2 * theta) is at its maximum value, which is 1.
* This happens when 2 * theta = 90 degrees, or theta = 45 degrees.

**5. Calculate the maximum distance:**

* Substitute theta = 45 degrees into the range equation:
   x_max = (v^2 * sin(90)) / g
   x_max = (v^2) / g

**Therefore, the maximum distance (range) of the projectile is achieved when the launch angle is 45 degrees, and the maximum distance is (v^2) / g.**