91Ó°ÊÓ

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 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

Understanding the Problem

We need to find out two things: (a) the maximum height that the helicopter reaches, given it has an upward acceleration and then enters free fall, and (b) the distance Powers is above the ground when the helicopter crashes, considering Powers uses a jetpack.

02

Calculate Helicopter Maximum Height

To find the maximum height reached by the helicopter, we first calculate the final velocity after 10 seconds of upward motion with acceleration 5.0 m/s². Using the formula \( v = u + at \), where initial velocity \( u = 0 \), \( a = 5.0 \text{ m/s}^2 \), and \( t = 10 \text{ s} \), we get \( v = 50 \text{ m/s} \). Next, use the formula \( s = ut + \frac{1}{2}at^2 \) to find the height after 10 seconds of ascent: \( s = \frac{1}{2} \times 5.0 \times 10^2 = 250 \text{ m} \).

03

Calculate Time to Reach Maximum Height from Free Fall

Once the engine is off, the helicopter moves under the influence of gravity alone. The initial velocity at engine cutoff is \( 50 \text{ m/s} \) upwards. To find the time to reach maximum height, set \( v = 0 \) in \( v = u - gt \), with \( g = 9.8 \text{ m/s}^2 \). Solving, \( 50 = 9.8t \) gives \( t \approx 5.10 \text{ s} \).

04

Calculate Additional Height During Free Fall Ascent

Using \( s = ut - \frac{1}{2}gt^2 \) for the additional height in free fall:\( s = 50(5.10) - \frac{1}{2} \times 9.8 \times 5.10^2 \approx 127.55 \text{ m} \). Total maximum height is the sum from powered ascent and free fall ascent: \( 250 \text{ m} + 127.55 \text{ m} \approx 377.55 \text{ m} \).

05

Calculate Time for Helicopter Descent

The helicopter starts descending after reaching the maximum height, taking time \( t \approx 9.81 \text{ s} \) to fall from 377.55 m to the ground using \(-377.55 = 0 - \frac{1}{2}gt^2 \).

06

Analyze Powers's Flight

Powers spends 7 seconds in free fall after leaving the helicopter at the velocity of 50 m/s upward. Use \( h = ut - \frac{1}{2}gt^2 \) for the height after 7 s: \( h = 50 \times 7 - \frac{1}{2} \times 9.8 \times 7^2 \approx 205 \text{ m} \).

07

Powers's Descent After Jet Pack Activates

After 7 seconds, Powers uses a jetpack, decelerating at 2 m/s². Calculate the new velocity: \( v = 50 - 9.8 \times 7 \approx -18.6 \text{ m/s} \). Another 4.10 seconds to land, using \( s = -18.6t + \frac{1}{2}\times 2 \times t^2 \). Total path is approximately 131.4 m from 205 m, remaining 73.6 m above ground.

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Key Concepts

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

Constant Acceleration

When an object experiences constant acceleration, it means the rate at which its velocity changes is consistent over time. In the context of our helicopter scenario, it has a constant upward acceleration of 5.0 m/s² as it initially takes off.

Why focus on constant acceleration? Because it provides a predictable method for calculating velocity and displacement over time.

• Using the formula \( v = u + at \), where \( u \) is the initial velocity, \( a \) is acceleration, and \( t \) is time, you can determine the velocity of an object after a specific period.
• By knowing how fast an object is moving, you can use other kinematic equations like \( s = ut + \frac{1}{2}at^2 \) to find out how far it has traveled.

In our exercise, the helicopter started from rest, so its initial velocity \( u \) was 0. After 10 seconds with a 5.0 m/s² acceleration, the helicopter gains a speed of 50 m/s. This is essential for understanding how high it travels before it starts to fall.

Free Fall

Free fall is a special condition where the only force acting on an object is gravity. In the problem, once the helicopter engine is turned off, the helicopter (as well as Agent Powers initially) is in free fall.

Why is free fall significant? Because it allows us to predict motion using gravity alone. Earth's gravity accelerates objects downward at 9.8 m/s², unless other forces like air resistance or propulsion intervene.

• In free fall, the upward motion of the helicopter after engine shutdown is countered by gravity. You can calculate how much higher it ascends before coming to a brief stop and then falling back down.
• To find out how long it takes to reach a peak (zero velocity), you adjust the velocity equation to \( v = u - gt \), solving for the time \( t \).

For our helicopter, this resulted in an additional 127.55 meters gained after engine cutoff, showing how significant free fall calculations are in understanding motion.

Jet Pack Mechanics

Jet packs provide controlled propulsion, allowing individuals like Agent Powers to manage their descent in a situation like this exercise.

When Powers activates his jet pack, he counteracts the free fall with a downward acceleration of 2.0 m/s². This presents an interesting mechanics problem:

• The jet pack reduces his downward velocity, allowing him more control over his landing.
• This change in acceleration needs to be calculated to determine how far Powers falls and his final state above the ground.

After turning on the jet pack, despite his initial speed being against him, Powers' controlled acceleration makes his descent safer and significantly slower than free fall would have been. This part of the problem teaches us the balance between powered versus uncontrolled motion.

Displacement Calculation

In kinematics, displacement is about finding out how far an object has moved in a given time frame. Displacement doesn’t just mean travel; it combines direction and distance.

Here's how displacement came into play in this exercise:

• Firstly, during the helicopter's powered ascent: Using \( s = ut + \frac{1}{2}at^2 \), the helicopter reached 250 meters.
• Next, the free-fall contribution: An additional 127.55 meters was achieved as the engine cut off.
• Lastly, Powers' jet pack reduces the net fall distance, controlling where he ends up relative to the crash site.

Understanding these calculations requires a solid grasp of initial conditions (like initial velocity) and how different accelerations change the motion over time. It shows clearly how, even with identical starting points, variations in journey (like the jet pack) affect ultimate displacement.