Consider a projectile of mass m which is shot vertically upward from the surface of the earth with initial velocity V. Assume that the gravitational force acts downward at a constant acceleration g while the force of air resistance has a magnitude proportional to the square of the velocity with proportionality constant k>0 and acts to resist motion. Let x=x(t) denote the height of the projectile at time t and v(t) = dx/dt(t) , its velocity.
a) Explain why the governing equation of motion is given by:
mdv/dt = -kv^2 - mg v > 0 For t > 0 (1)
mdv/dt = kv^2 - mg v < 0
x(0) = 0 and v(0) = Vo
b) Solve this system as follows: Introduce V(x) = v[t(x)]. Then define V1(x) = V^2(x) for V(x) >0 and V2(x) = V^2(x) for V(x) < 0. Show that V1 and V2 satisfy
dV1/dx + 2k/m V1 = -2g , V1(0) = Vo^2
dV2/dx - 2k/m V1 = -2g , V2(Xm) = 0
Where Xm is defined implicitly by V1(Xm) = 0. Demonstrate that V2(0) < Vo^2.© BrainMass Inc. brainmass.com July 15, 2018, 3:25 pm ad1c9bdddf
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v > 0 denotes the motion of the projectile upward. During the upward motion, forces on the projectile are: Downward gravitational force (mg) and the downward force of air resistance (kv^2).
Using Newton's second law in the upward direction,
m d^x/dt^2 = -kv^2 - mg
But, dx/dt = v
Hence, d^x/dt^2 = dv/dt
m dv/dt = -kv^2 - mg ------------- (1)
v < 0 denotes the ...
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