Problems for
Intermediate Methods in Theoretical Physics

Edward F. Redish

Time of Flight: 2

In a previous problem, we derived the equation for the time of flight, tF, of a projectile launched with a velocity v, launched at an angle θ from the vertical when viscous drag was included. The result was the equation

where

v0y = v0 cosθ = the y-component of the initial velocity

vT = mg/b = the terminal velocity

γ = b/m = the damping rate.

(a) Consider how you might solve this equation numerically (if you had values for the constants) by considering it as the intersection of two function graphs:

f(tF) = F(tF)

where

f(tF) = gtF

F(tF) = (v0y + vT)(1 - etF).

Sketch graphs for these two functions as a function of tF.

(b) The solution is at the intersection point of the two graphs. Since f is linear and the Taylor series for F shows that it starts of linearly and then slows its rate of increase, there is a possibility that the graphs will never cross (except at the origin). Let's consider for what values of the parameters there will be a non-zero solution. Assume that the angle is between 0 and 90o and that all of the constants are positive. First discuss from a physical point of view and then demonstrate your result mathematically.

(c) Expand F to second order in γtF ( = tF/tb, therefore a dimensionless number) and solve for tF.

(d) Show that if b = 0 you get the known result for time of flight, tF = 2v0y/g.

(e) Notice that if you put back the original constants (b, m, g) your solution is not second order in b. Arrange your constants so that b is grouped with other constants to make a dimensionless number and find the correction to tF to first order in this dimensionless number.


This page prepared by

Edward F. Redish
Department of Physics
University of Maryland
College Park, MD 20742
Phone: (301) 405-6120
Email: redish@umd.edu

Last revision 12. October, 2004.