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Chapter 2: Problem 52

A package is dropped from a helicopter that is descending steadily at a speed \(v_{0}\). After \(t\) seconds have elapsed, (a) what is the speed of the package in terms of \(v_{0}, g\), and \(t\) ? (b) What distance \(d\) is it from the helicopter in terms of \(g\) and \(t\) ? (c) What are the answers to parts (a) and (b) if the helicopter is rising steadily at the same speed?

Short Answer

Expert verified
For a descending helicopter, the speed of the package after \(t\) seconds is \(v_{0} + gt\) and the distance from the helicopter is \(v_{0}t + \frac{1}{2}g t^{2}\). If the helicopter is ascending, the speed of the package is \(-v_{0} + gt\) and the distance from the helicopter is \(-v_{0}t + \frac{1}{2}g t^{2}\).

Step by step solution

01

Determine the speed of the package

Firstly, determine the speed of the package after \(t\) seconds when the helicopter is descending at speed \(v_{0}\). The package will have an initial speed \(v_{0}\) (the same as the helicopter's). Under the influence of gravity, its speed after t seconds will be \(v_{0}\) (initial speed) + \(gt\) (increase in speed due to gravity, where \(g\) is the acceleration due to gravity). Therefore, the speed of the package \(v\) = \(v_{0} + gt\). Regarding sign, if downward is taken as positive, both \(v_{0}\) and \(gt\) will be positive.
02

Calculate the distance from the helicopter

Next, calculate the distance \(d\) the package is from the helicopter. The package falls under gravity, and hence its distance fallen under gravity after \(t\) seconds will be \(\frac{1}{2}g t^{2}\). However, the helicopter is also descending at speed \(v_{0}\), which increases the relative distance between them by \(v_{0}t\). Therefore, the total distance \(d\) = \(v_{0}t + \frac{1}{2}g t^{2}\).
03

Determine speed and distance for an ascending helicopter

Now, if the helicopter is ascending at speed \(v_{0}\), the package's initial speed will be \(-v_{0}\) (opposite in direction as compared to when the helicopter was descending). After \(t\) seconds, the speed of the package will be \(-v_{0} + gt\) (since \(gt\) is always positive as it is downward). The distance fallen under gravity will still be \(\frac{1}{2}g t^{2}\). However, as the helicopter is ascending, the relative distance between them will decrease by \(v_{0}t\). Therefore, the total distance \(d\) = \(-v_{0}t + \frac{1}{2}g t^{2}\).

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Key Concepts

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

Kinematics
When examining motion, kinematics is one of the most fundamental concepts. Kinematics focuses on describing how objects move without considering the forces that cause this movement. It involves parameters like displacement, velocity, acceleration, and time.

In the context of our exercise, we're concerned with the speed and distance a package travels after being dropped from a helicopter. Speed is how fast the package is moving, while distance measures how far it has traveled. By using kinematic equations, we can calculate these values given the initial speed and the acceleration due to gravity. These calculations are essential for understanding motion in one-dimensional situations, like vertical free fall.
Free fall
Free fall describes the motion of an object solely under the influence of gravity. In reality, this means no other forces, such as air resistance, significantly impact the motion.

For the falling package, free fall comes into play as soon as it鈥檚 released. With a constant gravitational pull, the package鈥檚 velocity changes steadily. The speed of the package increases by approximately 9.8 meters per second each second, determined by the acceleration due to gravity, symbolized as \(g\).
  • Initial speed in free fall: In our scenario, this could be either the descending or ascending speed of the helicopter.
  • Impact of gravity: As time increases, gravity works to accelerate the package toward the ground.
Relative motion
Relative motion looks at how the position of one object changes in relation to another moving object. Understanding this concept allows us to better interpret the observed changes in motion.

In our problem, we consider the package's motion relative to both the ground and the helicopter. For example, if the helicopter is moving upwards, the relative velocity of the package is different than if it were moving downwards. Therefore, the total distance includes what the helicopter traverses, impacting the distance calculated from the starting point.
  • Descending helicopter: Package's speed increases with both its initial speed and gravitational acceleration.
  • Ascending helicopter: Initially moves the package upward before gravity takes over, changing the relative distance.
Acceleration due to gravity
The acceleration due to gravity is a crucial constant in physics, often denoted by \(g\). On Earth, it is approximately \(9.8\, \/ \text{m/s}^2\), meaning that a freely falling object's speed increases by \(9.8\, \/ \text{m/s}\) every second.

In the problem of the falling package, \(g\) is the key factor that causes the package to accelerate downwards once it is released. We often use it in equations to calculate both the final speed and the distance traveled during free fall. By setting a direction as positive (usually downward), calculations become consistent and more manageable.
  • Equation for speed: \(v = v_0 + gt\)
  • Equation for distance fallen: \(d = \frac{1}{2}gt^2\)

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Most popular questions from this chapter

A car starts from rest and travels for \(t_{1}\) seconds with a uniform acceleration \(a_{1} .\) The driver then applies the brakes, causing a uniform acceleration \(a_{2}\). If the brakes are applied for \(t_{2}\) seconds, (a) how fast is the cas going just before the beginning of the braking period: (b) How far does the car go before the driver begins to brake? (c) Using the answers to parts (a) and (b) as the initial velocity and position for the motion of the ca during braking, what total distance does the car travel Answers are in terms of the variables \(a_{1}, a_{2}, t_{1}\), and \(t_{2}\).

A car accelerates uniformly from rest to a speed of \(40.0 \mathrm{mi} / \mathrm{h}\) in \(12.0 \mathrm{~s}\). Find (a) the distance the car travels during this time and (b) the constant acceleration of the car.

One athlete in a race running on a long, straight track with a constant speed \(v_{1}\) is a distance \(d\) behind a second athlete running with a constant speed \(v_{2}\). (a) Under what circumstances is the first athlete able to overtake the second athlete? (b) Find the time \(t\) it takes the first athlete to overtake the second athlete, in terms of \(d, v_{1}\), and \(v_{2}\). (c) At what minimum distance \(d_{2}\) from the leading athlete must the finish line be located so that the trailing athlete can at least tie for first place? Express \(d_{2}\) in terms of \(d, v_{1}\), and \(v_{2}\) by using the result of part (b).

In Bosnia, the ultimate test of a young man's courage used to be to jump off a 400 -year-old bridge (destroyed in 1993; rebuilt in 2004) into the River Neretva, \(23 \mathrm{~m}\) below the bridge. (a) How long did the jump last? (b) How fast was the jumper traveling upon impact with the river? (c) If the speed of sound in air is \(340 \mathrm{~m} / \mathrm{s}\), how long after the jumper took off did a spectator on the bridge hear the splash?

Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than \(800 \mathrm{~m} / \mathrm{s}^{2}\) lasting for any length of time will not cause injury, whereas an acceleration greater than \(1000 \mathrm{~m} / \mathrm{s}^{2}\) lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is \(0.40 \mathrm{~m}\) above the floor. If the floor is hardwood, the child's head is brought to rest in approximately \(2.0 \mathrm{~mm}\). If the floor is carpeted, this stopping distance is increased to about \(1.0 \mathrm{~cm}\). Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.
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