# Solution to equation of motion of a projectile shot upward

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.

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Answer:

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 ...

#### Solution Summary

This 4-page word document is full of explanations and detailed answer to the question of finding the equations of motion for an object shot vertically upward considering the effects of air resistance. This document also contains the solution process of the first order differential equation using a substitution. Equations are typed using the equation editor. Answer is very well formulated and presented.