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Chapter 2: Problem 74

A pilot flies horizontally at \(1300 \mathrm{~km} / \mathrm{h}\), at height \(h=35 \mathrm{~m}\) above initially level ground. However, at time \(t=0\), the pilot begins to fly over ground sloping upward at angle \(\theta=4.3^{\circ}\) (Fig. 2-38). If the pilot does not change the airplane's heading, at what time \(t\) does the plane strike the ground?

Short Answer

Expert verified
The plane strikes the ground at approximately 1.31 seconds.

Step by step solution

01

Understand the problem

We have a plane flying horizontally over level ground at a speed of \(1300 \text{ km/h}\), and it starts to fly over a slope going upwards at an angle of \(4.3^\circ\). Our task is to calculate the time \(t\) when the plane hits the ground, given that it maintains its initial horizontal trajectory, starting from a height of \(35 \text{ m}\).
02

Convert speed to meters per second

The speed of the plane is given as \(1300 \text{ km/h}\). We need to convert this speed to meters per second (m/s) for our calculations. \[1300 \text{ km/h} = 1300 \times \frac{1000}{3600} \approx 361.11 \text{ m/s}\]
03

Analyze the geometry of the problem

The ground is sloping upward, forming a right triangle. Initially, the plane is at \(35 \text{ m}\) height, and the slope rises at an angle \(\theta = 4.3^\circ\). We need to find the point where the vertical height of the plane becomes zero, indicating it has hit the ground.
04

Model the slope as a function

The slope of the ground forms a line with the equation relative to the horizontal flight. The height of the slope rises at a rate of \(\tan(4.3^\circ)\) for every meter the plane travels horizontally. Therefore, the height of the sloped ground at time \(t\) can be expressed as:\[ h_{\text{ground}} = v \cdot t \cdot \tan(4.3^\circ) \]where \(v\) is the speed of the plane in \(m/s\).
05

Determine time when plane hits the ground

The plane strikes the ground when its flying height equals the height of the sloped ground below. Set the initial height of the plane equal to the height of the slope:\[ 35 \text{ m} = 361.11 \cdot t \cdot \tan(4.3^\circ) \]Now solve for \(t\):\[ t = \frac{35}{361.11 \cdot \tan(4.3^\circ)} \]Calculate using a calculator:\[ \tan(4.3^\circ) \approx 0.0751\]\[ t \approx \frac{35}{361.11 \cdot 0.0751} \approx 1.309 \text{ seconds} \]
06

Solution verification

Verify the calculations in the equation for correctness. Substitute \(t\) back into the function:\[ 35 \approx 361.11 \cdot 1.309 \cdot 0.0751 \approx 35 \]The values match, confirming that the calculations are correct.

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Key Concepts

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

Velocity Conversion
When solving problems in kinematics, it's important to use consistent units. In this exercise, the airplane's velocity is given as 1300 km/h. To work with this speed in a standard SI unit of meters per second (m/s), we reference a crucial conversion factor. Dividing by 3.6 (because 1 km/h equals roughly 0.278 m/s) converts velocity from km/h to m/s.
The calculation is as follows:
  • Multiply the value in km/h by \( \frac{1000}{3600} \).
Using this conversion:
  • \( 1300 \text{ km/h} \times \frac{1000}{3600} \approx 361.11 \text{ m/s} \)
So, the plane's speed is about 361.11 m/s, which is necessary for further calculations in the problem. Always remember to convert measurements to appropriate units before performing any calculations involving motion.
Right Triangle
Understanding the formation of a right triangle can be critical in solving kinematic problems. In our exercise, as the plane travels over an inclined terrain, we picture a right triangle being formed between the horizontal ground, the slope (hypotenuse), and the vertical height between the plane and the inclined ground.
The role of the right triangle here is significant:
  • The horizontal distance the plane travels is one leg of the triangle.
  • The height change along the slope forms the other leg of the triangle.
  • The slope, inclined at a certain angle, is the hypotenuse.
Using the properties of the right triangle will help us determine how the vertical drop affects when the plane will touch the ground.
Trigonometry
Trigonometry plays a pivotal role in calculating the time before the airplane hits the ground. For any right triangle, the tangent of an angle can be defined as the ratio of the opposite side to the adjacent side. Here, the slope angle is \(4.3^\circ\).
To solve this problem:
  • We use the tangent function to calculate how quickly the elevation of the ground rises relative to the horizontal distance the plane travels.
  • \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \) translates to \( \tan(4.3^\circ) \) giving the slope rise rate.
By calculating \( \tan(4.3^\circ) \approx 0.0751 \), we can then find how fast the height increases per meter of travel, an essential step in determining when the plane meets the slope.
Slope Angle
The slope angle \((\theta)\) is crucial in understanding the plane's impending collision course with the ground. A slope angle signifies the angle at which land rises from a flat horizontal plane.
Key aspects of the slope angle include:
  • The steeper the slope, the faster the height rises as you move away horizontally.
  • In this problem, \( \theta = 4.3^\circ \), relatively mild, but enough to intercept the airplane's horizontal path given its height.
  • This angle enables the construction of the equation \( h_{\text{ground}} = v \cdot t \cdot \tan(4.3^\circ) \), relating time, slope height, and travel distance.
Understanding the influence of the slope angle helps predict when the plane will make contact with the inclined surface, considering it maintains a constant trajectory over a rising terrain.

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Most popular questions from this chapter

A ball of moist clay falls \(15.0 \mathrm{~m}\) to the ground. It is in contact with the ground for \(20.0 \mathrm{~ms}\) before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?

A lead ball is dropped in a lake from a diving board \(5.20 \mathrm{~m}\) above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. It reaches the bottom \(4.80 \mathrm{~s}\) after it is dropped. (a) How deep is the lake? What are the (b) magnitude and (c) direction (up or down) of the average velocity of the ball for the entire fall? Suppose that all the water is drained from the lake. The ball is now thrown from the diving board so that it again reaches the bottom in \(4.80 \mathrm{~s}\). What are the (d) magnitude and (e) direction of the initial velocity of the ball?

A red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an \(x\) axis, At time \(t=0\), the red car is at \(x_{r}=0\) and the green car is at \(x_{g}=\) \(220 \mathrm{~m}\). If the red car has a constant velocity of \(20 \mathrm{~km} / \mathrm{h}\), the cars pass each other at \(x=44.5 \mathrm{~m}\), and if it has a constant velocity of \(40 \mathrm{~km} / \mathrm{h}\), they pass each other at \(x=76.6 \mathrm{~m}\). What are (a) the initial velocity and (b) the constant acceleration of the green car?

The sport with the fastest moving ball is jai alai, where measured speeds have reached \(303 \mathrm{~km} / \mathrm{h}\). If a professional jai alai player faces a ball at that speed and involuntarily blinks, he blacks out the scene for \(100 \mathrm{~ms}\). How far does the ball move during the blackout?

A car moving with constant acceleration covered the distance between two points \(60.0 \mathrm{~m}\) apart in \(6.00 \mathrm{~s}\). Its speed as it passed the second point was \(15.0 \mathrm{~m} / \mathrm{s}\). (a) What was the speed at the first point? (b) What was the magnitude of the acceleration? (c) At what prior distance from the first point was the car at rest? (d) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the car, from rest \((t=0)\).
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