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Physics 101
Fall 2012
NAME:
Air Resistance Extra Credit
The following is worth 5 EXTRA CREDIT points!
We have seen that, in the absence of air resistance, a projectile travels along a parabolic
path given by
g
x2 ,
(1)
y(x) = tan θ x − 2
2v0 cos2 θ
where θ is the launch angle, g is the usual acceleration due to gravity, and v0 is the initial
launch speed. We now want to include the affects of air resistance and see what effect this
has on the projectile’s motion. Suppose that the projectile is subject to a drag force, F~D .
If the velocity of the projectile is not too high, then this drag force is proportional to it’s
velocity,
F~D = −b~v ,
(2)
where b is the proportionality constant which depends on the area of the projectile, density
of the air, etc. The only other force acting on the projectile is gravity, and so the force can
be resolved into components (here m is the mass of the projectile)
Fx = −bvx
Fy = −bvy − mg.
(3)
Newton tells us that F~ = m~a, and so dividing the terms in Eq. (3) by the mass gives
(defining γ ≡ b/m)
ax = v̇x = −γvx
(4)
ay = v̇y = −γvy − g,
where the dot denotes the derivative with respect to time, as always.
1. Assuming that the projectile starts with an initial velocity, v0x = v0 cos θ, along the x
direction, and v0y = v0 sin θ along the y direction, show that
vx (t) = v0 cos θe−γt
vy (t) =
− γg
+ v0 sin θ +
g
γ
(5)
e−γt
2. Using the fact that ex ≈ 1+x for small x, show that Eqs. (5) reduce to vx (t) = v0 cos θ,
and vy (t) = v0 sin θ − gt.
3. If the projectile starts from the origin, such that x0 = y0 = 0, show that
x(t) =
y(t) =
v0
cos θ (1 − e−γt )
γ
g
1
v0 sin θ + γ (1
γ
− e−γt ) −
gt
.
γ
(6)
4. Solve the first of Eqs. (6) to find t(x), and plug back into the second to determine
y(x). Show that your result reduces down to Eq. (1) when b = 0. Hint: you might
2
want to recall that ln (1 − x) ≈ −x − x2 for small x.
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