Problem 1 (Projectile motion with air resistance) We have seen in lecture that w

Question

Problem 1 (Projectile motion with air resistance) We have seen in lecture that
when an object is dropped from rest in the presence of both gravity and air resistance,
Newton’s second law of motion takes the following form:
y¨ = g +
c
m
y
2
, (1)
where y is the height of the object above the ground, g is the local acceleration due to gravity
near the earth’s surface, c is a constant, and m is the object’s mass. In this assignment, you
are going to find the analytical solution for the object’s trajectory y(t) subject to the initial
conditions
y(0) = y0, y(0) = 0.
(b) (5 points) Equation (1) appears to be a second-order ordinary differential equation.
However, it is really a first-order ODE in disguise. By defining a new variable v = y,
recast (1) as a first-order ODE in terms of v. What is the physical significance of v?
(c) (5 points) The “equilibrium” or “steady-state” value of v is the object’s terminal velocity
v. Based only on your answer to part (b), find v.
(d) (5 points) What is the initial condition for v, that is, v(0)?
(e) (15 points) By completing parts (b) and (d), you have reduced this second-order IVP
in y to a first-order IVP in v. By following the flowchart from lecture, solve for v(t).
Hint #1: You may find it useful to define a new (dimensionless!) variable u = v/v.
Hint #2: You will find the following identity useful:
Z
1
1 u
2
du = tanh1
(u) + c
(f) (5 points) Your expression for v(t) should involve the quantity gt/v. Show that this
quantity is simply t/ , where is the quantity you found in part (a) (iv).
(g) (5 points) is known as the characteristic time for this problem. Evaluate v( ), that
is, the velocity of the object at time t = . Based on your answer, what is the physical
significance of ?
(h) (5 points) Plot v/v versus t/ . Does your plot make sense physically?
(i) (10 points) Using your expression for v(t), solve for y(t).
Hint: You will find the following identity useful:
Z
tanh(u)du = ln [cosh (u)] + c
(j) (5 points) As a check on your work, make sure that your solution y(t) satisfies both the
governing ODE (1) and the boundary conditions (2).
(k) (5 points) Show that your expression for y(t) is consistent with the equation (4) you
derived in part (a) (v). Note that you may need to do a little algebraic manipulation.
(l) (5 points) Plot (c/m)y versus t/ for the initial condition y0=0. Does your plot make
sense physically?

Explanation / Answer

1. for a projectile motion with air resistance, the equation of motion of the particle becomes

y" = -g + cmy'^2

herer c is a constant and m is mass of the body

initial conditions

y(0) = yo

y'(0) = 0

b. now, v = y' ( speed of the particel)

then v' = -g + cmv ^2

here v is the speed of the particel

c. dv/dt = cm(-g/cm + v^2)

dv/(v^2 - g/cm) = cm*dt

integrating

(1/2*sqrt(g/cm))ln([v + sqrt(g/cm)]/[v - sqrt(g/cm)]) = cmt + K [ where K is constant of integration]

now at t = 0, v = 0

so,

(1/2*sqrt(g/cm))ln([ sqrt(g/cm)]/[ - sqrt(g/cm)]) = K = 0

hence

(1/2*sqrt(g/cm))ln([v + sqrt(g/cm)]/[v - sqrt(g/cm)]) = cmt

ln([v + sqrt(g/cm)]/[v - sqrt(g/cm)]) = 2t*sqrt(g*cm)

[v - sqrt(g/cm)]/[v + sqrt(g/cm)] = e^(-2t*sqrt(g*cm))

when t-> infinity

v = sqrt(g/cm)

d. intiial condition for v, v(0) = 0 ( because the ball is being dropped)

also, v(0) < sqrt(g/cm)

e. [v - sqrt(g/cm)]/[v + sqrt(g/cm)] = e^(-2t*sqrt(g*cm))

using componendo and dividendo

[v - sqrt(g/cm) + v + sqrt(g/cm)]/[v - sqrt(g/cm) - v - sqrt(g/cm)] = [e^(-2t*sqrt(g*cm)) + 1]/[e^(-2t*sqrt(g*cm)) - 1]

v = -sqrt(g/cm)[e^(-2t*sqrt(g*cm)) + 1]/[e^(-2t*sqrt(g*cm)) - 1]

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