91Ó°ÊÓ

Before understanding the parachutist's entire motion, we first need to grasp the concept of 'free fall'. Free fall occurs when an object moves under the force of gravity alone, with no other forces acting on it. For our parachutist, we ignore air resistance during this phase.

In free fall, the only force is gravity, which causes a constant acceleration of approximately 9.8 m/s\textsuperscript{2} (denoted as g). The displacement, or distance fallen, can be calculated using the kinematic equation: \ \( s = \frac{1}{2} g t^2 \).

Here, \( s \) is the distance (50 m), and \( t \) is the time taken to fall that distance. With g already known, we solve for time by rearranging the equation: \ \( t^2 = \frac{2s}{g} \).
Calculating this: \ \( t^2 = \frac{2 \times 50}{9.8} \approx 10.20 \) \ \( t \approx 3.2 \text{s} \).
Our parachutist falls for roughly 3.2 seconds before her parachute opens.

Once the parachute opens, the parachutist begins to decelerate. Deceleration means reducing the speed or negative acceleration. In this scenario, the deceleration rate given is \(2 \text{ m/s}^2 \).

This change in velocity is calculated using the formula: \ \( v_f = v_i + at \).
Here, \( v_f \) is the final velocity (3 m/s), \( v_i \) is the initial velocity (velocity at the end of free fall, which is 31.36 m/s), and \( a \) is the deceleration (-2 m/s\textsuperscript{2}). Solving for time \( t \), we rearrange the equation: \ \( t = \frac{v_f - v_i}{a} = \frac{3 - 31.36}{-2} \).
Calculating this gives us: \ \( t \approx 14.18 \text{s} \).
So, the deceleration lasts about 14.18 seconds.

The kinematic equations describe the motion of objects and are essential for these calculations.

Fundamental kinematic equations include: \ \( s = ut + \frac{1}{2} a t^2 \) \( (1) \)
\( v = u + at \) \( (2) \)
and \( v^2 = u^2 + 2as \) \( (3) \).

For our parachutist problem, we primarily use equations (1) and (2). Equation (1) helps determine the distance during deceleration: \ \( s = v_i t + \frac{1}{2} at^2 \).
Equation (2) calculates the final velocity after a certain time under uniform acceleration. It's used first to find the free-fall velocity: \ \( v = gt \).

Clearly understanding and manipulating these equations help solve parts (a) and (b) effectively.

Velocity calculation is essential in understanding motion. For a freely falling object, the velocity at any time can be found using: \ \( v = gt \).

In the parachutist problem, after 3.2 seconds of free fall, the velocity is: \ \( v = 9.8 \times 3.2 = 31.36 \text{ m/s} \).

With the parachute open, her velocity needed to change. We calculate the time it takes to decelerate to 3 m/s. Using the deceleration equation: \ \( 3 = 31.36 - 2t \),
we solve for \( t \) and find she decelerates over about 14.18 seconds. Calculations of velocities in different phases are vital for comprehending motion.

Calculating displacement (distance covered) helps understand the total fall height. During free fall: \ \( s = \frac{1}{2} g t^2 \).
Given \( s = 50 \) m for time \( t \approx 3.2 \) s.

During deceleration, we use: \ \( s = v_i t + \frac{1}{2} at^2 \).
Here, \( v_i = 31.36 \) m/s, \( a = -2 \) m/s\textsuperscript{2}, and \( t = 14.18 \) s.
Plugging in the values, the distance is: \ \( s = 31.36 \times 14.18 + \frac{1}{2} (-2) (14.18)^2 \approx 344.11 \text{ m} \).

Summing both displacements gives the total height from which she fell: \ \( h_{\text{total}} = 50 + 344.11 \approx 394.11 \text{ m} \).
Understanding displacement calculations clarify the complete motion.