CHARACTERISTICS of the PATH:

• The object follows the trace of a parabola.

• There is displacement in both the x and the y directions.

• The equation that describes the path is a quadratic or second degree equation.

  Hint: The x distance traveled along the path is the same each second. The y distance changes and is not the same!

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Equations: PLEASE NOTE that "V" is the launch velocity at angle q  from the horizontal

For x: The x position = (Vcosine q) * time = V t cos q

For y: The y position = (V sine q) *time - 1/2 g ²= V t sin q - 4.9 t ²

• Remember that velocity is a vector!

• The x component (of velocity "v") = v cosine q

• The y component (of velocity "v") = v sine q where q is the launch angle with the horizontal axis.

• Also remember to look at each dimension separately. Note that the x velocity is constant . This velocity moves the object forward at a constant rate. The y velocity goes from + v max (leaving launch level heading upward) to zero (top of path) to -vmax (back at launch level but heading downward).

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Vx = Vcos q

  Vy = V sin q + gt where g = - 9.8 m/s ² where V is the magnitude of launch velocity and q is the initial angle of launch

• In the x direction, a = 0 m/s^2 (if we ignore air resistance).
• The rate of change of the velocity is acceleration or a = Dv/Dt. For a projectile on the surface of the earth, a =g = - 9.8 m/s ² in the y direction only!

  ax =0

  Vx= V cos q

  Dx= (Vcosq)* t

ax = - 9.8 m/s^2

Vy= V sin q + gt

Dy= (Vsin q)* t + 1/2 gt ²

2aDy = V²y final - V²y initial      note: see above equations for finding Vy

4. The only   acceleration is  in the   y direction.

5. Only the y velocity is     zero at the top of the    path. The x velocity is    always there and is    constant!