This is a classical problem involving a parachutist falling under the influence of gravity and air resistance, and we need to use differential equations to solve it.

Step 1: Set up the differential equation when the parachute is closed

Initially, the parachutist is falling under the influence of two forces:

1. Gravity: $F_g = mg$, where $m = 70$ kg and $g = 9.81 \, \text{m/s}^2$.
2. Air resistance: $F_r = b_1 v$, where $b_1 = 30 \, \text{n-sec/m}$ and $v$ is the velocity.

The total force is then given by: $F = mg - b_1 v$ By Newton's second law, $F = ma$, where $a = \frac{dv}{dt}$, so: $m \frac{dv}{dt} = mg - b_1 v$ Substituting $m = 70$ and rearranging the equation: $\frac{dv}{dt} = g - \frac{b_1}{m} v$ Substitute $g = 9.81 \, \text{m/s}^2$ and $\frac{b_1}{m} = \frac{30}{70} \approx 0.4286$: $\frac{dv}{dt} = 9.81 - 0.4286 v$

Step 2: Solve the differential equation (closed parachute)

This is a first-order linear differential equation. The solution for $v(t)$ is: $v(t) = \frac{mg}{b_1} \left( 1 - e^{-\frac{b_1}{m}t} \right)$ Substitute $\frac{mg}{b_1} = \frac{70 \times 9.81}{30} \approx 22.87 \, \text{m/s}$: $v(t) = 22.87 \left( 1 - e^{-0.4286t} \right)$

Step 3: Find time when the velocity is 15 m/s (before the parachute opens)

We need to find $t_1$ when $v(t_1) = 15 \, \text{m/s}$. Set $v(t_1) = 15$: $15 = 22.87 \left( 1 - e^{-0.4286 t_1} \right)$ Solving for $e^{-0.4286 t_1}$: $\frac{15}{22.87} = 1 - e^{-0.4286 t_1}$ $e^{-0.4286 t_1} = 1 - 0.6558 = 0.3442$ Taking the natural logarithm on both sides: $-0.4286 t_1 = \ln(0.3442)$ $t_1 = \frac{\ln(0.3442)}{-0.4286} \approx 2.51 \, \text{seconds}$

So, after $t_1 = 2.51$ seconds, the parachute opens.

Step 4: Set up the differential equation when the parachute is open

Once the parachute opens, the air resistance changes to $b_2 = 90 \, \text{n-sec/m}$. The new differential equation is: $m \frac{dv}{dt} = mg - b_2 v$ $\frac{dv}{dt} = g - \frac{b_2}{m} v$ Substitute $\frac{b_2}{m} = \frac{90}{70} \approx 1.2857$: $\frac{dv}{dt} = 9.81 - 1.2857 v$

Step 5: Solve the differential equation (open parachute)

The solution to this equation is: $v(t) = \frac{mg}{b_2} \left( 1 - e^{-\frac{b_2}{m} (t - t_1)} \right)$ Substitute $\frac{mg}{b_2} = \frac{70 \times 9.81}{90} \approx 7.63 \, \text{m/s}$: $v(t) = 7.63 \left( 1 - e^{-1.2857(t - t_1)} \right)$ This is the velocity function for $t > t_1$.

Step 6: Calculate the time to reach the ground

We need to integrate the velocity to find the total distance traveled. The total distance is 2000 m, and the parachutist falls with two phases: before and after the parachute opens.

We can break the total fall distance into two parts:

• The distance fallen before the parachute opens, which we can calculate by integrating $v(t)$ from $0$ to $t_1$.
• The distance fallen after the parachute opens, which we can calculate by integrating $v(t)$ from $t_1$ to the time when the parachutist reaches the ground.

I will calculate these distances next.