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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

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?

Barry Bonds hits a home run so that the baseball just clears the top row of bleachers, \(21.0 \mathrm{m}\) high, located \(130 \mathrm{m}\) from home plate. The ball is hit at an angle of \(35.0^{\circ}\) to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time at which the ball reaches the cheap seats, and (c) the velocity components and the speed of the ball when it passes over the top row. Assume the ball is hit at a height of \(1.00 \mathrm{m}\) above the ground.

A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection?

A fisherman sets out upstream from Metaline Falls on the Pend Oreille River in northwestern Washington State. His small boat, powered by an outboard motor, travels at a constant speed \(v\) in still water. The water flows at a lower constant speed \(v_{w}\) He has traveled upstream for \(2.00 \mathrm{km}\) when his ice chest falls out of the boat. He notices that the chest is missing only after he has gone upstream for another 15.0 minutes. At that point he turns around and heads back downstream, all the time traveling at the same speed relative to the water. He catches up with the floating ice chest just as it is about to go over the falls at his starting point. How fast is the river flowing? Solve this problem in two ways. (a) First, use the Earth as a reference frame. With respect to the Earth, the boat travels upstream at speed \(v-v_{w}\) and downstream at \(v+v_{w \cdot}\) (b) \(\mathrm{A}\) second much simpler and more elegant solution is obtained by using the water as the reference frame. This approach has important applications in many more complicated problems; examples are calculating the motion of rockets and satellites and analyzing the scattering of subatomic particles from massive targets. PICTURE CANT COPY

A tire \(0.500 \mathrm{m}\) in radius rotates at a constant rate of 200 rev/min. Find the speed and acceleration of a small stone lodged in the tread of the tire (on its outer edge).

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