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When the Sun is directly overhead, a hawk dives toward the ground with a constant velocity of \(5.00 \mathrm{m} / \mathrm{s}\) at \(60.0^{\circ} \mathrm{be}-\) low the horizontal. Calculate the speed of her shadow on the level ground.

Short Answer

Expert verified
The speed of the hawk's shadow on the ground is equal to the horizontal component of the hawk's velocity, which is approximately \(2.5 \, m/s\).

Step by step solution

01

Identify Given Values and Required Value

The given values are the hawk's velocity \(5.00 \, m/s\) and the dive angle \(60.0^{\circ}\). The required value is the speed of the hawk's shadow, which represents the horizontal component of the hawk's velocity.
02

Determine the Relevant Component

Since the shadow's movement is along the ground, or horizontally, the speed of the shadow will be equivalent to the horizontal component of the hawk's velocity.
03

Utilization of Trigonometry to Find Horizontal Component

The horizontal component of a vector is given by the vector's magnitude times the cosine of its angle. Using this concept and the given values, the horizontal component of the hawk's velocity can be calculated as \(v_{horizontal}\) = \(v \cdot \cos(\theta)\), where \(v\) is the magnitude of the hawk's velocity and \(\theta\) is the dive angle. Substituting the given values, the horizontal component of the hawk's velocity becomes \(v_{horizontal}\) = \(5 \, m/s \cdot \cos(60.0^{\circ})\).

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Key Concepts

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

Projectile Motion
Projectile motion is a form of motion experienced by an object that is thrown near the Earth's surface and moves along a curved path under the action of gravity only. In our hawk example, when it dives towards the ground, we can view its motion as two independent components: one horizontal and one vertical. These two components are treated separately. The horizontal motion occurs at a constant speed because, in an ideal scenario without air resistance, there are no forces acting on the projectile horizontally once it's in motion. The vertical motion, on the other hand, is influenced by gravity, thus causing an acceleration.

The shadow of the hawk on the ground moves along with the horizontal component of the motion. This speed can be calculated separately from the vertical motion, which is precisely what makes understanding projectile motion so useful when trying to determine the speed of a moving shadow.
Vector Components
Vectors are mathematical entities that have both magnitude and direction. In physics, they are commonly used to represent quantities like velocity, force, and acceleration. The beauty of vectors lies in their ability to break down into components. By decomposing vectors into a horizontal (x-axis) and a vertical (y-axis) component, we simplify complex problems, making them easier to solve.

In the case of the diving hawk, we see vector components in action. The hawk's overall velocity is a vector that can be dissected into two perpendicular components: one representing the horizontal motion (towards the shadow) and the other, the vertical motion (towards Earth). Understanding how to work with vector components allows us to focus on the relevant dimension of the problem—in our case, the horizontal component that reveals the speed of the shadow on the ground.
Trigonometry in Physics
Trigonometry is an area of mathematics that deals with triangles, particularly right-angled triangles. It often plays a crucial role in physics when it comes to analyzing forces, motion, and vectors. The trigonometric functions—sine, cosine, and tangent—are especially useful when resolving a vector into its perpendicular components.

Considering the exercise of the hawk's dive, we've used the cosine function to figure out the horizontal velocity component from the overall velocity vector. This trigonometric function correlates the angle of the hawk's path with its horizontal speed. By multiplying the velocity magnitude by the cosine of the dive angle, we can isolate the speed of the shadow across the ground, which is essential to solving the problem at hand. Trigonometry, thus, provides the tools we need to translate angular information into linear speed—critical in our understanding of shadow motion.

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Most popular questions from this chapter

Note: Ignore air resistance in all problems and take \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) at the Earth's surface. To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of \(300 \mathrm{m} / \mathrm{s}\) at \(55.0^{\circ}\) above the horizontal. It explodes on the mountainside \(42.0 \mathrm{s}\) after firing. What are the \(x\) and \(y\) coordinates of the shell where it explodes, relative to its firing point?

A hawk is flying horizontally at \(10.0 \mathrm{m} / \mathrm{s}\) in a straight line, \(200 \mathrm{m}\) above the ground. A mouse it has been carrying struggles free from its grasp. The hawk continues on its path at the same speed for 2.00 seconds before attempting to retrieve its prey. To accomplish the retrieval, it dives in a straight line at constant speed and recaptures the mouse \(3.00 \mathrm{m}\) above the ground. (a) Assuming no air resistance, find the diving speed of the hawk. (b) What angle did the hawk make with the horizontal during its descent? (c) For how long did the mouse "enjoy" free fall?

The vector position of a particle varies in time according to the expression \(\mathbf{r}=\left(3.00 \hat{\mathbf{i}}-6.00 t^{2} \hat{\mathbf{j}}\right) \mathrm{m} .\) (a) Find expres- sions for the velocity and acceleration as functions of time. (b) Determine the particle's position and velocity at \(t=1.00 \mathrm{s}\)

The pilot of an airplane notes that the compass indicates a heading due west. The airplane's speed relative to the air is \(150 \mathrm{km} / \mathrm{h} .\) If there is a wind of \(30.0 \mathrm{km} / \mathrm{h}\) toward the north, find the velocity of the airplane relative to the ground.

A high-powered rifle fires a bullet with a muzzle speed of \(1.00 \mathrm{km} / \mathrm{s} .\) The gun is pointed horizontally at a large bull's eye target-a set of concentric rings- \(200 \mathrm{m}\) away. (a) How far below the extended axis of the rifle barrel does a bullet hit the target? The rifle is equipped with a telescopic sight. It is "sighted in" by adjusting the axis of the telescope so that it points precisely at the location where the bullet hits the target at \(200 \mathrm{m}\). (b) Find the angle between the telescope axis and the rifle barrel axis. When shooting at a target at a distance other than \(200 \mathrm{m}\) the marksman uses the telescopic sight, placing its crosshairs to "aim high" or "aim low" to compensate for the different range. Should she aim high or low, and approximately how far from the bull's eye, when the target is at a distance of (c) \(50.0 \mathrm{m},\) (d) \(150 \mathrm{m},\) or (e) \(250 \mathrm{m} ?\) Note: The trajectory of the bullet is everywhere so nearly horizontal that it is a good approximation to model the bullet as fired horizontally in each case. What if the target is uphill or downhill? (f) Suppose the target is \(200 \mathrm{m}\) away, but the sight line to the target is above the horizontal by \(30^{\circ} .\) Should the marksman aim high, low, or right on? (g) Suppose the target is downhill by \(30^{\circ} .\) Should the marksman aim high, low, or right on? Explain your answers.

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