Problem 100 - Fundamentals of Physics

Question

A parachutist bails out and freely falls \(50 \mathrm{~m}\). Then the parachute opens, and thercafter she decclerates at \(2.0 \mathrm{~m} / \mathrm{s}^{2} .\) She reaches the ground with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). (a) How long is the parachutist in the air? (b) At what height does the fall begin?

Step-by-Step Solution

Answer

Total time in air is sum of both phases; calculate initial fall height by combining sections.

Step 1: Calculate the Time for Free Fall

To solve this problem, we first determine the time the parachutist spends in free fall. The initial velocity during the free fall is 0 m/s. We use the formula for distance traveled during free fall: \[ s = \frac{1}{2}gt^2 \]where \( s = 50 \) m and \( g = 9.8 \) m/s². Solving for \( t \), we have: \[ 50 = \frac{1}{2} \times 9.8 \times t^2 \]\[ t^2 = \frac{50 \times 2}{9.8} \]\[ t^2 = \frac{100}{9.8} \approx 10.2 \]\[ t = \sqrt{10.2} \approx 3.2 \text{ seconds} \]. This is the time spent in free fall.

Step 2: Calculate the Speed Just Before Parachute Opens

Next, we calculate the speed of the parachutist just before the parachute opens using the formula:\[ v = gt \]where \( t = 3.2 \) seconds and \( g = 9.8 \) m/s²:\[ v = 9.8 \times 3.2 \approx 31.36 \text{ m/s} \].

Key Concepts

Free Fall

When talking about a parachutist in motion, one of the most interesting aspects is the period known as free fall. Free fall is the motion of a body subject only to gravity, without any air resistance or other forces acting. This means that the only force is the gravitational pull, which on Earth is about 9.8 m/s².
In the problem, the parachutist falls 50 meters freely. We can use a fundamental equation of kinematics to understand this motion:

• The equation \[ s = \frac{1}{2}gt^2 \]represents the distance \( s \) covered under constant acceleration.
• The initial speed \( v_0 \) is zero because the parachutist begins from rest.

Using this equation, we find the time spent in free fall, by applying the given distance and gravitational pull. This helps in understanding how long the parachutist traveled without any slowing forces acting, revealing the inherent simplicity and beautiful predictability of falling motion under gravity.

Deceleration

Once the parachute opens, the story changes from free fall to deceleration. Deceleration is simply negative acceleration, meaning the parachutist is slowing down. In the given problem, this deceleration is 2.0 m/s². Once her parachute opens, the force of air resistance begins to counteract the force of gravity, resulting in a net upward force that reduces her velocity.
This slowing effect increases the time it takes to descend the remaining distance to the ground safely. Since she reaches the ground with a final speed of 3.0 m/s, calculating the effects of constant deceleration is accomplished through the kinematics equations we have:

• Final velocity \( v = u + at \), and
• Displacement \( s = ut + \frac{1}{2}at^2 \), where \( u \) is the initial velocity after free fall.

The deceleration phase in parachuting is crucial for safety, showcasing how physics principles are applied in real-world scenarios to achieve safe landings.

Kinematics Calculations

Understanding parachutist motion also requires mastering kinematics calculations. These calculations help us understand how different phases of motion connect.

• The free fall relies on simple equations because there's only gravity acting.
• Upon parachute deployment, we incorporate the deceleration to model the change in motion.

By breaking the motion into different phases, the calculations help to predict how long the parachutist will remain in the air and at what initial height the jump begins. Applying kinematic equations strategically allows us to solve for variables like time, distance, and velocity in each phase. Knowing how and when to use these equations in parachuting and other real-life scenarios emphasizes their power and importance in physics.