Derive the trajectory equation by showing every step

Answer 1

To derive the trajectory equation, we can start with the basic kinematic equations of motion:

1. Position equation: x = x0 + v0*t + (1/2)*a*t^2
2. Velocity equation: v = v0 + a*t
3. Acceleration equation: v^2 = v0^2 + 2*a*(x - x0)

Let's assume we have a projectile motion, where an object is launched with an initial velocity v0 at an angle θ to the horizontal. The gravitational acceleration in the vertical direction is -g.

1. In the horizontal direction, there is no acceleration (assuming a vacuum), so the velocity and position equations become:
- x = x0 + v0*cos(θ)*t
- vx = v0*cos(θ)

2. In the vertical direction, the acceleration is equal to the gravitational acceleration, so the velocity and position equations become:
- y = y0 + v0*sin(θ)*t + (1/2)*(-g)*t^2
- vy = v0*sin(θ) - g*t

Now, let's eliminate the time variable to derive the trajectory equation. We can achieve this by combining the horizontal and vertical equations:

1. From the horizontal equation, we can solve for t: t = x / (v0*cos(θ))

2. Substitute this value of t into the vertical equation:
y = y0 + (v0*sin(θ)/cos(θ))*x/v0*cos(θ) + (1/2)*(-g)*(x^2/(v0^2*cos^2(θ)))
y = y0 + x*tan(θ) - (g*x^2)/(2*v0^2*cos^2(θ))

This is the equation of the projectile's trajectory. It relates the horizontal distance x to the vertical position y. The trajectory equation takes into account the initial height y0, initial velocity v0, launch angle θ, and the gravitational acceleration g.