Question 1209422: The parametric equations
x = vt \cos \theta,
y = vt \sin \theta - \frac{1}{2} gt^2,
describe the path of a projectile fired at an angle of theta, where v is the initial velocity, g denotes acceleration due to gravity, and t denotes time.
The path itself is a parabolic arch, and the distance the projectile reaches varies with the angle theta. Find the maximum distance.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **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.**
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