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Chapter 2: Problem 100

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?

Short Answer

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

Step by step solution

01

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/s虏. 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.
02

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/s虏:\[ v = 9.8 \times 3.2 \approx 31.36 \text{ m/s} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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/s虏.
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/s虏. 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.

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Most popular questions from this chapter

A car can be braked to a stop from the autobahn-like speed of \(200 \mathrm{~km} / \mathrm{h}\) in \(170 \mathrm{~m}\). Assuming the acceleration is constant, Find its magnitude in (a) SI units and (b) in terms of g. (c) How much time \(T_{b}\) is required for the braking? Your reaction time \(T,\) is the time you require to perceive an emergency, move your foot to the brake. and begin the braking. If \(T_{r}=400 \mathrm{~ms}\), then (d) what is \(T_{b}\) in terms of \(T_{n}\) and \((\mathrm{c})\) is most of the full time required to stop spent in reacting or braking" Dark sunglasses delay the visual signals sent from the cyes to the visual cortex in the brain, increasing \(T_{r-}\) (f) In the extreme case in which \(T_{r}\) is increased by \(100 \mathrm{~ms}\), how much farther does the car travel during your reaction time?

The position of a particle moving along an \(x\) axis is given by \(x=12 t^{2}-2 t^{3}\), where \(x\) is in meters and \(t\) is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at \(t=3.0 \mathrm{~s}\). (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and \(\underline{(g)}\) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at \(t=0\) )? (i) Determine the average velocity of the particle between \(t=0\) and \(t=\overline{3} \mathrm{~s}\)

A car traveling \(56.0 \mathrm{~km} / \mathrm{h}\) is \(24.0 \mathrm{~m}\) from a barrier when the driver slams on the brakes. The car hits the barrier \(2.00 \mathrm{~s}\) later. (a) What is the magnitude of the car's constant acceleration before impact? (b) How fast is the car traveling at impact?

A hot-air balloon is ascending at the rate of \(12 \mathrm{~m} / \mathrm{s}\) and is \(80 \mathrm{~m}\) above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what spced docs it hit the ground?

Two particles move along an \(x\) axis. The position of particle 1 is given by \(x=6.00 t^{2}+3.00 t+2.00\) (in meters and seconds); the acceleration of particle 2 is given by \(a=-8.00 t\) (in meters per second squared and seconds) and, at \(t=0,\) its velocity is \(20 \mathrm{~m} / \mathrm{s}\). When the velocities of the particles match, what is their velocity?
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