Here is my problem:

The equation: $\displaystyle m\frac{dv}{dt}=gm-Kv^2$

describes the velocity v of a parachute jumper. We have $\displaystyle v(0)=32,m=128, g=10$ and $\displaystyle K=5$.

1. Solve. I found $\displaystyle v(t)=\frac{16(3\exp(\frac{5}{4}t)+1)}{3\exp(\frac{ 5}{4}t)-1}$.

2. Find v(t) as $\displaystyle t \rightarrow \infty$. I found $\displaystyle v(t) \rightarrow 16$.

3. When is dv/dt the largest? At that moment, how many G's of force does the jumper experience? It is largest at t=0 at by replace the value of v at t=0 in the expression of dv/dt, I find dv/dt=-30. Therefore the jumper experiences 3G of force.

4. Here I am having trouble. If an auxiliary parachute opens at time $\displaystyle T$, the coefficient K=K(t) is $\displaystyle K(t)=5, 0 \leq t \leq T; K(t)=K_1, T<t$.

Without solving this new equation, figure out a reasonable choice for the constant $\displaystyle K_1$ so that the parachutist is not subject to worse $G$ forces that already exerted, but at the same time allows for the softest landing. With this choice of $\displaystyle K_1$, find the landing speed v (again without solving the equations).

Thank you in advance