Physics 374
Fall 2004
Prof. E. F. Redish

HW4, Problem 2

(a) 
m((d  (Overscript[v,  ->]) )/dt)=m(Overscript[g,  ->])  - b  (Overscript[v,  ->]) 

(b)
	m((Subscript[dv, x])/dt)=-b(Subscript[v, x])

	m((Subscript[dv, y])/dt)= - mg -b(Subscript[v, y])

Notice that (Overscript[g,  ->])  is a vector pointing down but g is a pure positive number.

(c) We didn't actually create a dimensionless equation last week, so let's just use the natural quantities we created.

	γ = (1/(Subscript[t, b])) = (m/b) (the damping rate)
\(TraditionalForm\`\(\(	\)\(v\_t\)\)\) = (mg/b)  (the terminal velocity)

((Subscript[dv, x])/dt)= - γ(Subscript[v, x])

solution is an exponential

(Subscript[v, x])(t) = (Ae^(-γt))

This satisfies the I.C.  (Subscript[v, x])(0) = (Subscript[v, 0]) cos θ = (Subscript[v, 0  x]) so 

(Subscript[v, x])(t) = (Subscript[v, 0  x]) (e^(-γt))

The equation for (Subscript[v, y]) is

	\(TraditionalForm\`dv\_y\/dt = \(\(-g\) - γv\_y = \(\(-\(γ(g\/γ + v\_y)\)\) = \(-\(γ(
                v\_t + v\_y)\)\)\)\)\)

so
	((Subscript[dv, y])/((Subscript[v, t]) + (Subscript[v, y]))) = -γdt
Solving
	ln((Subscript[v, t]) + (Subscript[v, y])) = -γt + C
Exponentiating
	(Subscript[v, t]) + (Subscript[v, y]) = (Ae^(-γt))
	(Subscript[v, y]) = (Ae^(-γt)) - (Subscript[v, t])
The I. C. is
	(Subscript[v, y])(
          0) = (Subscript[v, 0]) sin θ  =  (Subscript[v, 0  y]) =  
            A  -  (Subscript[v, t])
so
	A = (Subscript[v, t]) +  (Subscript[v, 0  y])	
and the solution is
	(Subscript[v, y]) = (Ae^(-γt)) - 
            (Subscript[v, t]) =  ((Subscript[v, t]) + (Subscript[v, 0  y])) (e^(-γt))  -  
            (Subscript[v, t])

A form in which it is more obvious what's going on is:

	(Subscript[v, y]) =  (Subscript[v, 0  y]) (e^(-γt))  +  
          (Subscript[v, t]) (1 - (e^(-γt)))
	
The initial velocity decays away (with the rate gamma) and is replaced by the terminal velocity.

(d) To get how long it was in the air we have to actually solve the position equation for y.  It's not enough to have v.  If there were no drag, we could do it. We could simply say that it would hit the ground when the velocity was the opposite of what it went up with.  But we can't say that here.  For example, it might have gone upward with a velocity greater than the terminal velocity and slowed down to a stop at the top.  When it falls from the top it will never fall faster than the terminal velocity.  So we have to actually find the position and find out when it gets back to the same starting point.

Since the equations separate, we don't have to bother with the x equation (unless we want to find the range -- which we might at some point).  The equation for y is

(dy/dt) =  (Subscript[v, y])  =  
          (Ae^(-γt)) - (Subscript[v, t])	(this is the easiest form to use since it puts the same kind of time dependent pieces together)
We can read off the solution.

y(t) = 
        C - (1/γ) (Ae^(-γt)) - (Subscript[v, t]) t

The constant A was fixed in getting the initial velocity.  The constant C must be chosen to get the initial position.  Taking y(0) = 0 this gives

C = (1/γ) A

Putting everything together we get:

y(
          t) = (1/γ) A(1 - (e^(-γt))) - (Subscript[v, t]) t

(Check the value at t = 0.)  The projectile will come back to the ground when y(t)=0 again.  This means

(Subscript[v, t]) 
          (Subscript[t, F]) = (1/γ) A(1 - (e^(-γ(Subscript[t, F]))))

It's straightforward to put back the values for A, γ, and (Subscript[v, t]) to get the desired result.

(Solution by E. F. Redish)

(Solution by E .  Redish)