Notes In Applied Mathematics: 1 y186o

Math 312 - Applied Math handout # 1 Vertical projectile motion with resistance proportional to speed Consider Newton’s Second Law for vertical motion (with y measured upwards): dv dt

k mv

g, v 0

v0

(1)

Equation (1) can be solved using the integrating factor exp kt/m or by the method of undetermined coefficients (try both if you’ve taken an ODE class!); the latter is a little quicker, so I write ce

v and using the initial condition, we find that c vt

gm k

kt/m

gm/k

v 0 and so the speed of the ball is given by

gm e k

v0

gm k

kt/m

(2)

Notice from this equation that the terminal speed v is defined by v

gm k

lim v t t

which, for a feather, snowflake or whiffle ball, is pretty much achieved before it hits the ground. Suppose however that we are really only interested in the altitude y as a function of time; OK – if the ball is projected upward from y 0 then y 0 0. Hence from (2) yt

t 0

v0

gm e k

ks/m

gm ds k

(3)

where s is a dummy variable for t. Upon integrating we obtain yt

m v 0 gm e ks/m gms | s t s 0 k k k gm gm m v0 1 e kt/m t, 0 k k k

(4) t

T

where T is the time at which the ball hits the ground. Now for an interesting question: does the ball take longer to rise to its maximum height, or to fall back from that height? All I’m going to do here is to graph y t for various values of k/m and v 0 , and then draw some conclusions. Choosing k/m

6 s 1 and v 0

5, 10, 15, 20 m/s, with g yt

1. 1056 1

e

6t

9. 8 m/s 2 we find for v 0

5 m/s

1. 6333t

the graph of which is the lowest of the four shown; the others get higher with increasing values of v 0 as would be expected. Note that the graph below is not the path of the projectile.

3 2.5 2 1.5 1 0.5 0

0.2 0.4 0.6 0.8 1

Now what do you think?

1.2t 1.4 1.6 1.8 2

2.2 2.4