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Chapter 2: Problem 63

A helicopter 8.50 m above the ground and descending at 3.50 m/s drops a package from rest (relative to the helicopter). Just as it hits the ground, find (a) the velocity of the package relative to the helicopter and (b) the velocity of the helicopter relative to the package. The package falls freely.

Short Answer

Expert verified
(a) 13.38 m/s downward (b) -13.38 m/s upward

Step by step solution

01

Understand the given data

The helicopter is at a height of 8.50 m above the ground and is descending at a velocity of 3.50 m/s. The package is dropped from rest relative to the helicopter, and we need to find its velocity just before it hits the ground. The acceleration due to gravity is approximately \( g = 9.81 \, \text{m/s}^2 \).
02

Identify the motion of the package

The package starts from rest relative to the helicopter, which means its initial velocity relative to the ground is the same as the helicopter: -3.50 m/s (descending). As it falls freely, it accelerates under gravity.
03

Use the kinematic equation for velocity

To find the velocity of the package when it hits the ground, we use the kinematic equation: \[ v^2 = u^2 + 2as \] where \( v \) is the final velocity, \( u \) is the initial velocity (-3.50 m/s), \( a \) is the acceleration due to gravity (9.81 m/s虏), and \( s \) is the distance fallen (8.50 m).
04

Calculate the final velocity of the package

Substitute the known values into the equation: \[ v^2 = (-3.50)^2 + 2 \times 9.81 \times 8.50 \] \[ v^2 = 12.25 + 166.77 \] \[ v^2 = 179.02 \] Taking the square root, we get:\[ v = \sqrt{179.02} \approx 13.38 \, \text{m/s} \]Thus, the final velocity of the package relative to the ground is approximately 13.38 m/s downward.
05

Find the velocity of the package relative to the helicopter

The package's velocity relative to the helicopter is the same as its velocity relative to the ground because the initial velocity of the package relative to the helicopter was 0.
06

Find the velocity of the helicopter relative to the package

The velocity of the helicopter relative to the package is the negative of the velocity of the package relative to the helicopter, which is \(-13.38 \, \text{m/s}\) upward.

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Key Concepts

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

Velocity
Velocity is a key concept in kinematics, describing the speed and direction of an object's movement. In the context of this exercise, we need to determine the velocity of both the package and the helicopter.

When the package is dropped, it does not start from complete rest because it inherits the downward velocity of the helicopter. Therefore, its initial velocity relative to the ground is -3.50 m/s, indicating that it is moving downwards at the same rate as the helicopter.

The kinematic equations, such as the equation to find velocity, are crucial here. The equation \[ v^2 = u^2 + 2as \] helps determine how the velocity of an object changes when subjected to constant acceleration. Knowing the initial velocity (\(u = -3.50 ext{ m/s}\)) and calculating the final velocity (\(v\)) can help us understand the movement dynamics of the package just before impact.

Remember:
  • The velocity of the helicopter is constant during its descent, contributing to the initial velocity of the package.
  • Negative signs in velocity indicate downward movement, which is significant for understanding direction.
Free Fall
Free fall refers to the motion of an object where gravity is the only force acting on it. In our exercise, once the package is released from the helicopter, it enters a state of free fall. It means the package is influenced solely by gravity, without other forces like air resistance affecting it (in ideal conditions).

The motion in free fall is characterized by the acceleration due to gravity, \(g\), which is approximately \(9.81 ext{ m/s}^2\) on Earth. This acceleration is constant and acts downwards.

Free fall allows us to simplify calculations for kinematics, using formulas like \[ v = u + at \] to find velocity after a certain time or \[ v^2 = u^2 + 2as \] to find the velocity knowing the distance covered.

Key points to remember:
  • Objects in free fall experience a uniform acceleration due to gravity.
  • The absence of air resistance in ideal calculations simplifies the mathematical treatment of motion.
  • Initial conditions, such as starting velocity, influence the outcome of free fall calculations.
Acceleration due to Gravity
Acceleration due to gravity is a fundamental component of Earth's environment that influences how objects fall. Represented by \(g\), its standard value is \(9.81 ext{ m/s}^2\). This universal constant denotes how fast velocities change when in free fall.

In any descent, such as our package from the helicopter, gravity exerts a downward pull, increasing the package's velocity as it falls. Regardless of the object's mass, gravity causes all objects to accelerate at the same rate toward Earth.

To apply this concept:
  • Use \(g\) to calculate velocity changes using equations of motion (e.g., \(v = u + gt\)).
  • Remember that free fall involves only gravity, making calculations dependent solely on this constant and initial conditions.
  • Consider the effect of gravity in more complex scenarios such as projectile motion or orbiting bodies.

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Most popular questions from this chapter

(a) The pilot of a jet fighter will black out at an acceleration greater than approximately 5\(g\) if it lasts for more than a few seconds. Express this acceleration in \(\mathrm{m} / \mathrm{s}^{2}\) and \(\mathrm{ft} / \mathrm{s}^{2}\) (b) The acceleration of the passenger during a car crash with an air bag is about 60\(g\) for a very short time. What is this acceleration in \(\mathrm{m} / \mathrm{s}^{2}\) and \(\mathrm{ft} / \mathrm{s}^{2}\) (c) The acceleration of a falling body on our moon is 1.67 \(\mathrm{m} / \mathrm{s}^{2} .\) How many \(g^{\prime}\) is this? (d) If the acceleration of a test plane is \(24.3 \mathrm{m} / \mathrm{s}^{2},\) how many \(g^{\prime}\) is it?

A Toyota Prius driving north at 65 \(\mathrm{mph}\) and a \(\mathrm{VW}\) Passat driving south at 42 \(\mathrm{mph}\) are on the same road heading toward each other (but in different lanes). What is the velocity of each car relative to the other (a) when they are 250 \(\mathrm{ft}\) apart, just before they meet, and (b) when they are 525 \(\mathrm{ft}\) apart, after they have passed each other?

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Look out below. Sam heaves a 16 -lb shot straight upward, giving it a constant upward acceleration from rest of 45.0 \(\mathrm{m} / \mathrm{s}^{2}\) for 64.0 \(\mathrm{cm} .\) He releases it 2.20 \(\mathrm{m}\) above the ground. You may ignore air resistance. (a) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?
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