Analyze the flight of a projectile moving under the influence of gravity (mg) and linear, i.e., laminar flow drag (fD = bv, with b > 0 constant). Neglect
Let us define projectile motion as the motion of a particle through a region of space where it is subject to constant acceleration. An object moving through the air near the surface of the earth is subject to the constant gravitational acceleration g, directed downward. If no other forces are acting on the object, i.e. if the object does not have a propulsion system and we neglect air resistance, then the motion of the object is projectile motion.
Assume that we want to describe the motion of such an object,
starting at time t = 0. Let us orient our coordinate system such
that one of the axes, say the yaxis, points upward. Then
a = a_{y}j =
gj, a_{y} = g. We can rotate our
coordinate system about the yaxis until the velocity vector of the
object at t = 0 lies in the xy plane, and we can choose the origin
of our coordinate system to be at the position of the object at t =
0.
Since we have motion with constant acceleration, the velocity at time t will be v = v_{0} + at, and the position will be r = r_{0} + v_{0}t + (1/2)at^{2}. Writing this equation in component form and using r_{0} = 0, v_{0} = v_{0x}i + v_{0y}j, and a = a_{y}j = gj we have v = v_{0x}i + v_{0y}j  gtj, r = v_{0x}ti + v_{0y}tj  (1/2)gt^{2}j. 

The position vector r and the velocity vector
v at time t have only components along the x and
yaxes. By choosing a convenient orientation of our coordinate
system we have simplified the mathematics involved in solving a
projectile motion problem. We can treat projectile motion
asmotion in two dimensions with
v_{x} = v_{0x}, x = x_{0} + v_{0x}t, v_{y}= v_{0y}  gt, y = y_{0} + v_{0y}t  (1/2)gt^{2}. If the initial velocity v_{0} makes an angle ?_{0} with the xaxis, then v_{0x} = v_{0}cos?_{0} and v_{0y} = v_{0}sin?_{0}. We then have v_{x} = v_{0}cos?_{0} = constant, x = x_{0} + v_{0}cos?_{0}t, v_{y} = v_{0}sin?_{0}  gt, y = y_{0} + v_{0}sin?_{0}t  (1/2)gt^{2}. We can solve x = v_{0}cos?_{0}t for t in terms of x, t = x/(v_{0}cos?_{0}) and substitute this expression for t into y = v_{0}sin?_{0}t  (1/2)gt^{2}. We obtain (y  y_{0}) = (x  x_{0})tan(?_{0})  g (x  x_{0})^{2}/(2v_{0}^{2}cos^{2}(?_{0})), an equation for the path (or trajectory) of the object 
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