Mechanics question with parachute

[Sink41]

1. A parachutist of mass m falls freely until his parachute opens. When it is open he experiences an upward resistance kv where v is his speed and k is a positive constant.

The parachutist falls from rest freely under gravity for a time $\frac{m}{2k}$ and then opens his parachute. Prove that the total distance he has fallen when his velocity is $\frac{3mg}{4k}$ is $\frac{m^2 g}{8 k^2} (8 \ln2 - 1)$

2.
$$a = v \frac{dv}{ds}$$

$$v^2 = u^2 + 2 a s$$

$$v = u + a t$$

3. Until parachute is opened:

$$t = \frac{m}{2k}$$

$$a = g$$

$$u = 0$$$$v = u + a t$$

$$v = \frac{mg}{2k}$$$$v^2 = u^2 + 2 a s$$

$$\frac{m^2 g^2}{4 k^2} = 2 g s$$

$$\frac{m^2 g}{8 k^2} = s$$

When parachute is opened:

Force downwards $ma = mg - kv$

$$m v \frac{dv}{ds} = mg - kv$$

$$\int \frac{mv}{mg - kv} dv = \int ds$$Substitution: $$u = mg - kv$$

$$\frac{du}{dv} = -k$$

$$v = \frac{u - mg}{-k}$$$$\int \frac{mv}{-k (mg - kv)} du = \int ds$$

$$\int \frac{m(u - mg)}{u} du = \int ds$$

$$\int m - \frac{m^2 g}{u} du = \int ds$$

$$mu - m^2 g \ln u = \int ds$$

$$m (mg - kv) - m^2 g \ln (mg - kv) = s + c$$ where c is a constant

To work out constant:

$$v = \frac{mg}{2k}$$

$$s = \frac{m^2 g}{8 k^2}$$$$m^2 g - mk \frac{mg}{2k} - m^2 g \ln ( mg - \frac{kmg}{2k} ) = \frac{m^2 g}{8 k^2} + c$$

$$\frac{m^2 g}{2} - m^2 g \ln ( \frac{mg}{2} ) - \frac{m^2 g}{8 k^2} = c$$

Putting c back into equation:

$$m^2 g - mkv - m^2 g \ln ( mg - kv ) - \frac{m^2 g}{2} + m^2 g \ln ( \frac{mg}{2} ) + \frac{m^2 g}{8 k^2} = s$$

Putting in value of v:

$$- \frac{m^2 g}{4} + m^2 g \ln 2 + \frac{m^2 g}{8 k^2} = s$$

$$\frac{m^2 g}{8} ( 8 \ln 2 - 2 + \frac{1}{k^2} ) = s$$So i have wrong answer where have i gone wrong?

 

[Fermat]

...

When parachute is opened:

Force downwards $ma = mg - kv$

$$m v \frac{dv}{ds} = mg - kv$$

$$\int \frac{mv}{mg - kv} dv = \int ds$$


Substitution: $$u = mg - kv$$

$$\frac{du}{dv} = -k$$

$$v = \frac{u - mg}{-k}$$


$$\int \frac{mv}{-k (mg - kv)} du = \int ds$$

$$\int \frac{m(u - mg)}{k^2u} du = \int ds$$



...

Your error is shown above in the last line. You canceled the (-k) on the bottom instead of getting (-k)² (on the bottom)

 

[funnyfox]

u have given time is m/2k and k is a constant(dimensionless)...so how come time have dimensions "kg" (m->kg)...so check ur qstn..

 

[Fermat]

k is a constant but it cerainly isn't dimensionless.

We are told: Restance = kv
and [R] = N

Since k = R/v, then

[k] = [N/(m/s)]=[Ns/m] - the dimensions of k.

Then [m/2k] = [kg/(Ns/m)] = [kgm/Ns] = - the dimension of time.