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Chapter 3: Problem 57

You are in a hot-air balloon that, relative to the ground, has a velocity of \(6.0 \mathrm{m} / \mathrm{s}\) in a direction due east. You see a hawk moving directly away from the balloon in a direction due north. The speed of the hawk relative to you is \(2.0 \mathrm{m} / \mathrm{s} .\) What are the magnitude and direction of the hawk's velocity relative to the ground? Express the directional angle relative to due east.

Short Answer

Expert verified
The hawk's velocity relative to the ground is \(6.32\,\mathrm{m/s}\) at \(18.4^{\circ}\) north of east.

Step by step solution

01

Define the Problem

We need to find the velocity of the hawk relative to the ground. The balloon's velocity relative to the ground is \(6.0 \; \mathrm{m/s}\) east. The hawk's velocity relative to the balloon is \(2.0 \; \mathrm{m/s}\) north. We'll find the vector sum of these velocities to get the hawk's velocity relative to the ground.
02

Set Up Velocity Vectors

The velocity of the balloon relative to the ground is \(\mathbf{v}_{b} = 6.0 \; \mathrm{m/s} \; \text{east}\). Write this as a vector: \(\mathbf{v}_b = 6.0 \hat{i}\). The hawk's velocity relative to the balloon is \(\mathbf{v}_{h/b} = 2.0 \; \mathrm{m/s} \; \text{north}\), which can be written as \(\mathbf{v}_{h/b} = 2.0 \hat{j}\).
03

Add Vectors to Find Hawk鈥檚 Velocity

The hawk's velocity relative to the ground \(\mathbf{v}_h\) is the sum of \(\mathbf{v}_b\) and \(\mathbf{v}_{h/b}\): \[ \mathbf{v}_h = \mathbf{v}_b + \mathbf{v}_{h/b} = 6.0 \hat{i} + 2.0 \hat{j} \]
04

Calculate Magnitude of Hawk鈥檚 Velocity

To find the magnitude of \(\mathbf{v}_h\), use the Pythagorean theorem:\[ |\mathbf{v}_h| = \sqrt{(6.0)^2 + (2.0)^2} = \sqrt{36 + 4} = \sqrt{40} = 6.32 \; \mathrm{m/s} \]
05

Calculate Directional Angle of Hawk鈥檚 Velocity

To find the angle \( \theta \) relative to due east, use the tangent function: \[ \tan \theta = \frac{2.0}{6.0} \]Solving for \( \theta \):\[ \theta = \tan^{-1}\left(\frac{2}{6}\right) = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.4^{\circ} \]
06

State Final Velocity of Hawk

The hawk has a velocity of \(6.32 \; \mathrm{m/s}\) at an angle of \(18.4^{\circ}\) north of east relative to the ground.

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Key Concepts

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

Vector Addition
In physics, vectors are used to represent quantities that have both magnitude and direction, such as velocity. When adding vectors, you essentially combine these quantities based on their respective dimensions. This is done to find a resultant vector, which represents the cumulative effect.

To add two vectors, one representing the velocity of the hot air balloon and another for the hawk relative to the balloon, we align them in a coordinate system:
  • The balloon's velocity is directed entirely east, represented by vector \(6.0 \hat{i}\).
  • The velocity of the hawk relative to the balloon is north, given by vector \(2.0 \hat{j}\).
The vector sum, therefore, is the combination of both directional velocities:\[ \mathbf{v}_h = 6.0 \hat{i} + 2.0 \hat{j} \]This process of addition tells us the overall velocity of the hawk relative to the ground.
Pythagorean Theorem
When dealing with vectors on a plane, the Pythagorean theorem is incredibly useful to determine the magnitude of the resultant vector. This method is applied when the vectors are perpendicular to each other, such as our east-west and north-south vectors.

The formula for the Pythagorean theorem is \( a^2 + b^2 = c^2 \). In our scenario:
  • The velocity in the east direction is \(6.0 \; \mathrm{m/s}\) \((a = 6.0)\).
  • The velocity in the north direction is \(2.0 \; \mathrm{m/s}\) \((b = 2.0)\).
Using these values, we calculate the magnitude of the hawk's velocity relative to the ground:\[ \sqrt{6.0^2 + 2.0^2} = \sqrt{36 + 4} = \sqrt{40} = 6.32 \; \mathrm{m/s} \]This tells us how fast the hawk is moving in total.
Velocity Components
Velocity components break down the movement of an object into defined directions, often using a coordinate system like Cartesian coordinates. Here, the components are crucial for understanding how the hawk travels concerning the ground.

  • East Component: This is contributed entirely by the velocity of the balloon, which is \(6.0 \; \mathrm{m/s}\) along the x-axis.
  • North Component: Directly contributes \(2.0 \; \mathrm{m/s}\) along the y-axis, representing the movement of the hawk relative to the balloon.
By utilizing these components, we can represent complex movements with simpler vector additions, making it easier to calculate the resultant velocity relative to the ground.
Directional Angle
Determining the directional angle of a vector is critical for fully describing its orientation. In this context, the directional angle relates to how the hawk's trajectory veers from due east.

To find this angle, we use the tangent function, as it links the ratios of the opposite side to the adjacent side in a right triangle. Here
  • The opposite side is the northward velocity of \(2.0 \; \mathrm{m/s}\).
  • The adjacent side is the eastward velocity of \(6.0 \; \mathrm{m/s}\).
The formula used is:\[ \tan \theta = \frac{2.0}{6.0} \]Solving for the angle \( \theta \) gives us:\[ \theta = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.4^{\circ} \]This angle indicates that the hawk's path is \(18.4^{\circ}\) north of east, providing a complete understanding of its direction.

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Most popular questions from this chapter

A fire hose ejects a stream of water at an angle of \(35.0^{\circ}\) above the horizontal. The water leaves the nozzle with a speed of \(25.0 \mathrm{m} / \mathrm{s}\). Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.050 s, during which time it experiences an acceleration of \(340 \mathrm{m} / \mathrm{s}^{2} .\) The ball is launched at an angle of \(51^{\circ}\) above the ground. Determine the horizontal and vertical components of the launch velocity.

A Motorcycle Jump. You are planning to make a jump with your motorcycle by driving over a ramp that will launch you at an angle of \(30.0^{\circ}\) with respect to the horizontal. The front edge of the ramp on which you are supposed to land, however, is 25.0 ft lower than the edge of the launch ramp (i.e., your launch height). (a) Assuming a launch speed of \(65 \mathrm{mph}\), at what horizontal distance from your launch point should the landing ramp be placed? (b) In order to land smoothly, the angle of the landing ramp should match the direction of your velocity vector when you touch down. What should be the angle of the landing ramp?

A swimmer, capable of swimming at a speed of \(1.4 \mathrm{m} / \mathrm{s}\) in still water (i.e., the swimmer can swim with a speed of \(1.4 \mathrm{m} / \mathrm{s}\) relative to the water), starts to swim directly across a \(2.8-\mathrm{km}\) -wide river. However, the current is \(0.91 \mathrm{m} / \mathrm{s},\) and it carries the swimmer downstream. (a) How long does it take the swimmer to cross the river? (b) How far downstream will the swimmer be upon reaching the other side of the river?

A projectile is launched from and returns to ground level, as the figure shows. There is no air resistance. The horizontal range of the projectile is measured to be \(R=175 \mathrm{m},\) and the horizontal component of the launch velocity is \(v_{0 x}=+25 \mathrm{m} / \mathrm{s} .\) Concepts: (i) What is the final value of the horizontal component \(v_{x}\) of the projectile's velocity? (ii) Can the time be determined for the horizontal part of the motion? (iii) Is the time for the horizontal part of the motion the same as that for the vertical part? (iv) For the vertical part of the motion, what is the displacement of the projectile? Calculations: Find the vertical component \(v_{0 y}\) of the projectile.
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