We are given the following information:

• The parachutist falls 40 m before the parachute opens.
• After the parachute opens, the retardation (deceleration) is 2 m/s².
• The parachutist reaches the ground with a velocity of 3 m/s.
• Gravitational acceleration $g = 10 \, \text{m/s}^2$.

We need to determine:

1. The height from which the parachutist jumped (i.e., the height of the airplane).
2. The total time the parachutist remains in the air.

Step 1: Find velocity when the parachutist falls the first 40 m

For the first part of the fall (before the parachute opens), the parachutist is in free fall. We can use the following kinematic equation to find the velocity after falling 40 m:

$v^2 = u^2 + 2 g h$ where:

• $u = 0 \, \text{m/s}$ (initial velocity),
• $g = 10 \, \text{m/s}^2$,
• $h = 40 \, \text{m}$.

$v^2 = 0 + 2 \times 10 \times 40 = 800 \implies v = \sqrt{800} = 28.28 \, \text{m/s}.$

So, after falling 40 m, the parachutist has a velocity of $28.28 \, \text{m/s}$.

Step 2: Find the distance traveled after the parachute opens

When the parachute opens, the parachutist experiences a retardation of $2 \, \text{m/s}^2$ and reaches the ground with a velocity of $3 \, \text{m/s}$. Using the kinematic equation:

$v_f^2 = v_i^2 + 2 a d$ where:

• $v_f = 3 \, \text{m/s}$ (final velocity),
• $v_i = 28.28 \, \text{m/s}$ (initial velocity when the parachute opens),
• $a = -2 \, \text{m/s}^2$ (retardation),
• $d = ?$ (distance after the parachute opens).

Solving for $d$:

$3^2 = 28.28^2 + 2 \times (-2) \times d$ $9 = 799.94 - 4d$ $4d = 799.94 - 9 = 790.94$ $d = \frac{790.94}{4} = 197.735 \, \text{m}.$

So, the parachutist falls an additional $197.74 \, \text{m}$ after the parachute opens.

Step 3: Calculate the height of the airplane

The total height from which the parachutist jumped is the sum of the two distances:

$\text{Total height} = 40 \, \text{m} + 197.74 \, \text{m} = 237.74 \, \text{m}.$

Step 4: Find the total time in the air

Time for the first part (free fall for 40 m):

We can use the equation for velocity:

$v = u + g t$ $28.28 = 0 + 10 t_1$ $t_1 = \frac{28.28}{10} = 2.828 \, \text{seconds}.$

Time for the second part (with parachute opened):

For the second part of the fall, we use the equation:

$v_f = v_i + a t$ $3 = 28.28 - 2 t_2$ $2 t_2 = 28.28 - 3 = 25.28$ $t_2 = \frac{25.28}{2} = 12.64 \, \text{seconds}.$

Total time:

$\text{Total time} = t_1 + t_2 = 2.828 + 12.64 = 15.468 \, \text{seconds}.$

Final Answers:

• The height of the airplane was approximately $237.74 \, \text{m}$.
• The parachutist remained in the air for approximately $15.47 \, \text{seconds}$.

Would you like more details on any specific part?

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Here are five related questions for further understanding:

1. How would the total time change if the parachutist fell from a higher altitude?
2. How does the retardation of the parachute affect the total time of descent?
3. What would happen to the velocity if the parachutist experienced no retardation?
4. How does the gravitational acceleration influence the velocity during free fall?
5. What happens if the parachute opens at a different height?

Tip: Always double-check unit consistency when applying physics formulas, as they are key to accurate calculations!