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A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2} .\) Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 \(\mathrm{s}\) , Powers shuts off the engine and steps out of the helicopter Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back 7.0 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0 \(\mathrm{m} / \mathrm{s}^{2} .\) How far is Powers above the ground when the helicopter crashes into the ground?

Step by step solution

01

Calculate Velocity at Shutdown

The helicopter has a constant upward acceleration of \(5.0 \, \text{m/s}^2\) for \(10.0\) seconds. Using the formula \( v = u + at \), where initial velocity \( u = 0 \), acceleration \( a = 5.0 \, \text{m/s}^2 \), and time \( t = 10 \, \text{s} \), the velocity at engine shutdown is \( v = 0 + 5.0 \times 10 = 50 \, \text{m/s} \).

02

Determine Maximum Height During Ascent

We need to calculate the distance traveled during the powered ascent. Use \( s = ut + \frac{1}{2}at^2 \) with \( u = 0 \), \( a = 5.0 \text{ m/s}^2 \), and \( t = 10 \text{ s} \). So, \( s = 0 + \frac{1}{2} \times 5.0 \times 10^2 = 250 \, \text{m} \).

03

Calculate Additional Height During Free Fall

During free fall, initial velocity \( v = 50 \, \text{m/s} \), downward acceleration due to gravity is \( g = -9.8 \, \text{m/s}^2 \). At maximum height, final velocity \( v = 0 \, \text{m/s} \). Use \( v^2 = u^2 + 2as \) to find additional height: \( 0 = 50^2 + 2(-9.8)s \), \( s = \frac{2500}{19.6} \approx 127.6 \, \text{m} \).

04

Total Maximum Height Above Ground

Add heights from powered ascent and free ascent: \( 250 + 127.6 = 377.6 \, \text{m} \). So, the maximum height of the helicopter is \( 377.6 \, \text{m} \).

05

Calculate Time to Fall from Maximum Height

Use \( v = u + at \) to find the time \ taken for the helicopter to hit the ground from maximum height, with initial velocity \( u = 0 \, \text{m/s} \), final velocity \( v \) is irrelevant when hitting the ground, and \( a = 9.8 \, \text{m/s}^2 \). Use \( s = 377.6 \) to solve \( s = ut + \frac{1}{2}at^2 \) for \( t \). This gives \( 377.6 = 0 + \frac{1}{2} \times 9.8 \times t^2 \). Solving gives \( t \approx 8.8 \, \text{s} \).

06

Calculate Distance Powers Travels After Deployment

Powers free falls for 7.0 seconds and then deploys a jet pack. In 7 seconds, he falls \( s = 50 \times 7 - \frac{1}{2} \times 9.8 \times 7^2 = 245 \, \text{m} \) below 377.6 m, leaving him at 132.6 m above the ground. After deployment, he travels for \( t = 1.8 \text{ s} \) at \( a = 2.0 \text{ m/s}^2 \) more: \( s = \frac{1}{2} \times 2 \times 1.8^2 = 3.24 \, \text{m} \) downward. Therefore, Powers is \(132.6 - 3.24 = 129.36 \, \text{m} \) above the ground.

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Key Concepts

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

Projectile Motion

Projectile motion describes the motion of an object that is launched into the air and continues its movement mainly under the influence of gravity. In this context, the helicopter carrying Dr. Evil is initially in projectile motion during the phase when Austin Powers deactivates the engine, and they both ascend into the air. This phase involves a trajectory that features both horizontal and vertical components, typical of projectile motion scenarios.

• During the initial ascent, powered by the helicopter's engines, both vertical velocity and distance increase.
• Once the engine is cut, the helicopter continues its ascent due to inertia, transitioning into free fall when it reaches its peak.

Understanding the helicopter's trajectory involves examining both the ascent under constant acceleration due to the engines and the subsequent descent governed by gravity alone.

Free Fall

Free fall is a specific type of projectile motion where the only force acting on an object is gravity. In the scenario, when Austin Powers turns off the helicopter's engine, it transitions into free fall.

• The helicopter begins to fall under the sole influence of Earth's gravity, accelerating down at approximately 9.8 meters per second squared.
• At this point, there are no other forces acting on the helicopter; the upward forces previously provided by the engine have ceased.
• Free fall continues until either Austin Powers activates his jet pack or another force acts on him or the helicopter.

By examining this phase, physics students can explore the fundamental principles of gravitational acceleration and how they apply in real-world situations.

Constant Acceleration

Constant acceleration refers to the steady increase in velocity, either due to a force or as a natural component like gravity. The helicopter's movement and Austin Powers' subsequent descent encompass two different phases of constant acceleration:

• Initially, the helicopter experiences a constant upward acceleration of 5.0 \(\text{m/s}^2\) while its engine is active.
• Once the engine is off, it begins to accelerate downward at a constant rate of 9.8 m/s², the acceleration of gravity.
• Later, when Austin Powers uses his jet pack, he decelerates at a constant rate of 2.0 m/s².

Each phase serves as an excellent example of constant acceleration's effects on velocity and displacement, which are crucial concepts for problem-solving in physics.

Physics Problem Solving

Physics problem solving involves breaking complex situations into solvable parts using known formulas and principles. The exercise with Austin Powers and the helicopter is an exercise-rich environment for practicing these skills:

• Start by identifying all forces and types of motion involved — initially the powered ascent, then free fall, and finally the controlled jet pack descent.
• Apply appropriate formulas like \( v = u + at \) for computing velocity during acceleration phases and \( s = ut + \frac{1}{2}at^2 \) for displacement.
• Combine calculations for each phase to get the complete picture of the motion — like finding total height using additive calculations of each segment's height.

This approach not only helps solve the given problem but also builds a strong foundation for tackling more complex physics challenges.