Answer:

Okay since offline po kayo, may dalawang cases po kasi ang projectile motion. And iba po an pagkuha niyan per case.

(note: I will use vi = initial velocity. and vf = final velocity. not sure if v0 and v1 po gamit niyo tho, also θ is theta which is yung angle)

Case 1: Thrown an an angle

Horizontal Dimension(x)

Initial Velocity

  vix = vi(cosθ)

Final Velocity

  vfx = vi(x)     (x is yung horizontal distance)

Acceleration

  ax = 0m/s²   (kasi constant po)

Time of flight

ang binigay sa amin na formula is time of flight from the ground to the maximum height. bale, mahati yung trajectory / parabola.

  t = -viy / ay  (kalahati ng parabola)

  So total time is 2t

    Also yung t na kalahati ang gagamitin pangsubstitute sa mga t dito sa baba

Displacement

Maximum Height

  h = viy(t) + 1/2(ay)(t²)

Range

   R = vix(t) or R = 2vix(-viy / ay)

Vertical Dimension (y)

Initial Velocity

viy = vi(sin θ)

Final Velocity

vfy = viy + gt

Acceleration

ay = -9.8 m/s²