PROJECTILE MOTION SIMULATION

The Calculations

Since we have horizontal acceleration as a function with respect to time we know acceleration is not constant and hence SUVAT equations do not hold.
Instead we have a differential equations with respect to time. We integrate at each timestep and find the constant to get velocity, then again to get displacement. Repeating this multiple times every second we get enough position points to plot the whole curve.
Note: each new position depends on the old position so there is no way to calculate the final position/time/velocity without calculating all the points inbetween.

The inital conditions for this projectile are given by:

Initial height: 0, Initial Vertical Velocity : 21 , Initial Horizontal Velocity : 21

Horizontal Velocity = $\int$-1 t² + 0 t + 2 dt = -0.3 t³ + 0t² + 2t + c₁

where $c$₁ is found by solving the equation with the initial conditions. Doing this we get $c$₁ = 21 . So Horizontal Velocity $\approx$22

Vertical Velocity = $\int \; -9.81\; dt\; =\; -9.81t\; +\; c$₂$$

where $c$₂ is found by solving the equation witht the initial conditions. Doing this we get $c$₂ = 21 So Vertical Velocity $\approx$17

Then we integrate again to get the displacement,

Horizontal Diplacement = $\int$-0.3t³ + 0 t² + 2 t + 21 dt = -0.1t⁴ + 0t³ + 1t² + 21t + c₃

where $c$₃ is found by solving the equation with the initial conditions. Doing this we get $c$₃=0 So Horizontal Displacement $\approx$10

Vertical Displacement = $\int \; -9.81t\; +$21 dt = 4.9t² + 21t + c₄

where $c$₄ is found by solving the equation with the given initial conditions. Doing this we get $c$₄ = 0 So Vertical Displacement $\approx$9

Note: as with all projectile motion examples we are making a lot of assumptions: (1) the vertical acceleration is constant and is given by gravity (2) horizontal acceleration is variable and can be given as a polynomial of degree at most 2 with respect to time (3) we are taking the earth as flat (3) each axis is a 'wall' (4) the particle does not bounce