Solve the differential equation.
A particle of mass m is thrown vertically upwards against gravity and is subjected to an air resistance where `k` is a constant and `v` is the velocity of the particle in the upward direction. Using Newton's law of motion the velocity of the particle satisfies the differential equation `v(dv)/(ds) = - (g+kv)`
where s is the distance travelled in the upward direction and `g` is the gravitational constant.
Given the initial condition that v = U when s = 0, solve the ODE to obtain the distance s in terms of v.
Hence obtain the formula for the max height reached by the particle.
Hint: Use partial fractions to evaluate the integral.
We start with the equation:
`v(dv)/(ds) = -g-kv`
Let's manipulate the equation to try to separate all v terms from all s terms:
`(dv)/(ds) = -(g+kv)/v`
`1/(ds) = -(g+kv)/(vdv)`
Well, we've separated the terms, let's just invert both fractions:
`ds = -(vdv)/(g+kv)`
Now, we do as we always do with separated variables: integrate!
`intds = -int(vdv)/(g+kv)`
The integral on the left is easy. To evaluate the integral on the right, we can integrate by rewriting the numerator and denominator and by multiplying (effectively) by k/k:
`-1/k int (kv +g - g)/(g+kv) dv = -1/k int 1 - g/(g+kv) dv`
Now, we have an evaluable integral! We get the following result:
`-1/k(v - g/kln|g+kv| + C)`
Now, we'll multiply through by (-1/k) to finish off our evaluation of the integral (keep in mind, `-1/kC` is just another constant, so let's just let that `C` absorb the -1/k):
`-v/k - g/k^2ln|g+kv| + C`
Now, we have our differential equation in an evaluable form! Also, we can get rid of the absolute value because we're looking only at upward velocities according to the problem.
`s+C_1 = -v/k-g/k^2ln(g+kv) + C`
Again, when we subtract the constant of integration from the left side from the constant of integration on the right side, we get just another random constant. We have now solved for `s` in terms of `v`:
`s = -v/k - g/k^2 ln(g+kv)+ C`
Now, we solve for the constant. Recognize that we are given an initial condition, when s = 0, v = U:
`0 = -U/k - g/k^2ln(g+kU) + C`
and our overall equation becomes:
`s(v) = -v/k - g/k^2ln(g+kv) + U/k + g/k^2ln(g+kU)`
We can simplify (based on log rules and the distributive property):
`s(v) = (U-v)/k + g/k^2ln((g+kU)/(g+kv))`
Alright, now that we have our equation, let's go ahead and determine the maximum height reached by the particle. When the particle has reached its maximum height, the velocity is zero, so let's solve for s given that v = 0.
`s(0) = U/k + g/k^2 ln((g+kU)/g)`
Can't really simplify this much more, so we're good!
Fun times! Hope that helped.