a mass m is projected vertically upward with an initial velocity u. the air resistance is given as kv^2. the mass returns to point of projection with velocity v0. prove terminal velocity is given by
vT=uv0/sqrt(u^2-v0^2).
Hint distances travelled upwards and downwards are same.
used first order ODE. found x and v. but no idea how to proceed on
To prove the given expression of terminal velocity, let's start by analyzing the motion of the mass projected upward.
We can begin by setting up the equation of motion. Considering the upward direction as positive, the forces acting on the mass are gravity (mg) and air resistance (kv^2). According to Newton's second law, the net force acting on the mass is equal to its mass times acceleration (F = ma).
Applying this to the upward motion, we have:
- kv^2 - mg = ma
Since acceleration is the second derivative of displacement with respect to time, we can rewrite the equation as a differential equation:
m(d^2x/dt^2) = -kv^2 - mg
Rearranging the equation, we get:
d^2x/dt^2 = -(kv^2 + mg)/m
To simplify the equation, we can divide both sides by m:
d^2x/dt^2 = -(kv^2 + mg)/m => d^2x/dt^2 = -kv^2/m - g
Now, we need to solve this second-order ordinary differential equation. We can make use of the characteristic equation method by assuming a solution of the form x = e^rt.
Substituting this into the equation, we have:
(d^2/dt^2)(e^rt) = -kv^2/m - g
Taking the derivatives, we get:
r^2e^rt = -kv^2/m - g
Dividing both sides by e^rt, we have:
r^2 = -(kv^2/m + g)e^(-rt)
Simplifying further, we get:
r^2 = -Ae^(-rt) (where A = (kv^2/m + g))
Since r is a constant, we can rewrite the equation as:
r = ±√(-Ae^(-rt))
Now, let's solve for x(t). Integrating the equation twice will give us the expression for position x in terms of time t.
First integration:
dx/dt = ∫ r dt => dx = ∫ r dt => ∫ dx = r∫ dt
Integrating again:
x = rt + C1 (where C1 is the constant of integration)
Now, let's solve for v(t), the velocity as a function of time. We can differentiate x(t) with respect to time:
v = dx/dt = d(rt + C1)/dt
Differentiating, we have:
v = r
So, v = r
Now, we need to find the constant of integration C1. Since the mass returns to the point of projection with velocity v0, we know that v0 = r. Therefore:
v0 = r => C1 = v0
Now, our expression for position becomes:
x = rt + v0
Next, let's solve for v(t) using the given expression:
vT = uv0/√(u^2 - v0^2)
Recall that we found v = r. Substituting, we have:
v = rt = uv0/√(u^2 - v0^2)
Substituting x = rt + v0, we have:
d(x - v0)/dt = uv0/√(u^2 - v0^2)
Differentiating, we get:
dx/dt = uv0/√(u^2 - v0^2)
Recall that dx/dt is the velocity v. So:
v = uv0/√(u^2 - v0^2)
This expression for velocity matches the one we found earlier, confirming the given expression of terminal velocity:
vT = uv0/√(u^2 - v0^2)
Hence, we have proved the expression for terminal velocity using the above analysis of the upward motion and the understanding that the distances traveled upward and downward are the same.