Analyze the flight of a projectile moving under the influence of gravity (mg) and linear, i.e.,...

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

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Answer #1

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 y-axis, points upward. Then a = ayj = -gj, ay = -g. We can rotate our coordinate system about the y-axis until the velocity vector of the object at t = 0 lies in the x-y 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 = v0 + at, and the position will be r = r0 + v0t + (1/2)at2. Writing this equation in component form and using r0 = 0, v0 = v0xi + v0yj, and a = ayj = -gj we have

v = v0xi + v0yj - gtj,     r = v0xti + v0ytj - (1/2)gt2j.

The position vector r and the velocity vector v at time t have only components along the x- and y-axes. 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

vx = v0x,    x = x0 + v0xt,

vy= v0y - gt,    y = y0 + v0yt - (1/2)gt2.

If the initial velocity v0 makes an angle ?0 with the x-axis, then v0x = v0cos?0 and v0y = v0sin?0. We then have

vx = v0cos?0 = constant,   x = x0 + v0cos?0t,

vy = v0sin?0 - gt,   y = y0 + v0sin?0t - (1/2)gt2.

We can solve x = v0cos?0t for t in terms of x, t = x/(v0cos?0) and substitute this expression for t into y = v0sin?0t - (1/2)gt2. We obtain

(y - y0) = (x - x0)tan(?0) - g (x - x0)2/(2v02cos2(?0)),

an equation for the path (or trajectory) of the object

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