# 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.

© BrainMass Inc. brainmass.com October 24, 2018, 10:57 pm ad1c9bdddfhttps://brainmass.com/math/calculus-and-analysis/solution-to-equation-of-motion-of-a-projectile-shot-upward-176436

#### Solution Preview

Following is the text part of the solution. Please see the attached file for complete solution. Equations, diagrams, graphs and special characters will not appear correctly here. Thank you for using Brainmass.

===============================================================================

(1)

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.

Explanation of Projectile Motion

Question: A projectile is shot from the edge of a cliff 285 m above ground level, with an initial speed of vo=135 m/s and at an angle of 37.0 degrees with the horizontal, as shown in the attached figure. Please view attachment.

a) Determine the time taken by the projectile to hit point P at ground level.

b) Determine the range X of the projectile as measured from the base of the cliff.

c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Assume the positive directions are upward and to the right.)

d) What is the magnitude of the velocity?

e) What is the angle made by the velocity vector with the horizontal?(below the horizontal)