Final answer:
A rocket launched from an 8-ft platform at a 60-degree angle with an initial velocity of 144 ft/s follows projectile motion. Its path is modeled by the equation y = 8 + 124.7t - 16t², which is a parabola.
Step-by-step explanation:
To find the rectangular equation that models the path of a rocket launched from an 8-ft platform with an initial velocity of 144 ft per sec at a 60-degree angle, we need to use the principles of projectile motion.
First, we split the initial velocity into its horizontal and vertical components. Since the angle given is 60 degrees and the initial velocity is 144 ft per sec, we can calculate the components as follows:
- Horizontal Component (Vx) = 144 * cos(60) = 144 * 0.5 = 72 ft/s
- Vertical Component (Vy) = 144 * sin(60) = 144 * (√3/2) ≈ 124.7 ft/s
The horizontal motion is at a constant velocity, so the horizontal position x at time t is given by x = Vx * t. The vertical motion is under gravity, starting with an initial velocity Vy, so the vertical position y at time t can be found using the equation y = Vy * t - (1/2) * g * t^2, where g is the acceleration due to gravity (32 ft/s²).
Because the rocket starts from an 8-ft platform, the equation for the vertical position y will have an extra term, becoming y = 8 + Vy * t - (1/2) * g * t^2.
The path of the rocket is a parabola, which is characteristic of projectile motion without any form of air resistance or other forces.
In summary, the rectangular equation that models the rocket's path is:
y = 8 + 124.7t - 16t²
This equation yields a parabolic path, which means the rocket will first ascend and then descend, following the shape of a parabola.