Question

A projectile is ground-launched into the air at time t = 0. The initial speed is vo and the angle to
the horizontal is ?. Air resistance is modeled as a drag force acting on the projectile and results
in an acceleration aD = ?kv where k is a constant and v is the velocity of the projectile. If the
effects of gravitational acceleration are included,

a) Determine expressions for the x- and y- components of the displacement of the
projectile as functions of time

Answer 1

the velocity at angle ? is u, then is the vertical component of the velocity
vv = u*sin? , and the horizontal velocity is
vh = u*cos?.

according to s = v*t, the horizontal displacement after time t is : x = vh*t = u*cos?*t
the vertical displacement y after time t is y = vv*t - 1/2*gt^2 = u*sin?*t - 0.5*gt^2.

o find the cartesian equation, eliminate t:
x = u*cos?*t---> t = x/(u*cos?) --> plug into equation for y:

y = u*sin?*(x/(u*cos?) - 1/2*g(x^2/(u^2*cos^2(?))
y = tan?*x - gx^2/(2u^2*cos^2(?)) = cartesian equation of trajectory.
If the starting point is at height h, the equation is
y = tan?*x - gx^2/(2u^2*cos^2(?)) + h
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to find the range set y = 0 , because when the projectile hits the ground at range x, y = 0:
0 = tan?*x - gx^2/(2u^2*cos^2(?))
0 = - sin?/cos?)*(2u^2*cos^2(?))/g*x + x^2
0 = - sin?*2u^2*cos?/g*x + x^2 -->
use the identity sin(2?) = 2sin?*cos?
0 = -sin(2?)*u^2/g*x + x^2
solve the quadratic function for x:
x = u^2*sin(2?)/g = range when start and end point are same level.

for the height set kinetic energy = potential energy:
the kinetic energy = 1/2*m*(vv)^2
the potential energy = mgh
1/2*(vv)^2 = gh --> solve for h:
h = (u*sin?)^2/(2g) = maximum height above start level.

For the range, when start- and end-point are not at the same level, use the
equation
y = tan?*x - gx^2/(2u^2cos^2(?)) + h and set y = 0, and solve for range x.