/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 A dolphin leaps out of the water... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 67

A dolphin leaps out of the water at an angle of \(35^{\circ}\) above the horizontal. The horizontal component of the dolphin's velocity is \(7.7 \mathrm{m} / \mathrm{s}\). Find the magnitude of the vertical component of the velocity.

Short Answer

Expert verified
The vertical component of the dolphin's velocity is approximately 5.39 m/s.

Step by step solution

01

Identify Known Values

The problem provides that the dolphin leaps at an angle of \(35^{\circ}\) above the horizontal and the horizontal component of the velocity is \(7.7\, \mathrm{m/s}\).
02

Use Trigonometric Relation to Find Vertical Component

Since we have the horizontal component (\(v_x\)) and the angle (\(\theta\)), we use the sine function to find the vertical component (\(v_y\)) of the velocity. The relation is \( \sin(\theta) = \frac{v_y}{v} \). But we want to express \(v_y\) in terms of \(v_x\), so the formula becomes \( v_y = v_x \cdot \tan(\theta) \).
03

Calculate the Vertical Component

Use the trigonometric identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \) and \( v_x = v \cdot \cos(\theta) \). Thus, \( v_y = v_x \cdot \tan(35^{\circ}) \). Substitute \( \tan(35^{\circ}) \approx 0.7002 \) and \( v_x = 7.7 \) to find \( v_y \approx 7.7 \times 0.7002 = 5.39 \).
04

Finalize the Answer

After calculating, the magnitude of the vertical component of the velocity is approximately \(5.39\, \text{m/s}\).

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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 fantastic example of the curved paths objects follow through the sky. When the dolphin leaps out of the water, it experiences projectile motion. This type of motion combines two separate components: horizontal and vertical. Each component acts independently of the other. Thus, the horizontal component moves at a constant velocity because no extra forces act upon it. Meanwhile, the vertical component is affected by gravity, causing the dolphin to follow a parabolic path. By understanding these two components, we can predict where an object will land and estimate its time in the air.
Velocity Components
In projectile motion, velocity is split into horizontal and vertical components. Understanding how to separate and calculate these components is crucial in physics.
When the dolphin jumps at an angle, its speed is divided into two parts:
  • The horizontal component ( \( v_x \)) stays constant, since no horizontal forces change during the motion.
  • The vertical component ( \( v_y \)) changes due to gravitational force acting downwards.
To find \( v_y \), we use the tangent of the angle, given by \( \tan(\theta) = \frac{v_y}{v_x} \). Therefore, \( v_y = v_x \cdot \tan(\theta) \). This allows us to find how the angle influences the vertical speed.
Physics Problem Solving
Solving physics problems involves logical thinking and applying the right equations correctly. It is essential to understand what the problem asks and identify the information given.
  • First, look for key variables provided, such as angles or known velocities.
  • Next, choose the appropriate trigonometric relations or formulas to relate these variables.
  • Finally, calculate carefully, considering units and rounding where necessary.
In our dolphin example, recognizing the trigonometric functions suitable for deriving the vertical velocity is crucial. Developing these skills enables students to tackle any problem and understand the movement and forces involved.

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Most popular questions from this chapter

A puck is moving on an air hockey table. Relative to an \(x, y\) coordinate system at time \(t=0\) s, the \(x\) components of the puck's initial velocity and acceleration are \(v_{0 x}=+1.0 \mathrm{m} / \mathrm{s}\) and \(a_{x}=+2.0 \mathrm{m} / \mathrm{s}^{2} .\) The \(y\) components of the puck's initial velocity and acceleration are \(v_{0 y}=+2.0 \mathrm{m} / \mathrm{s}\) and \(a_{y}=-2.0 \mathrm{m} / \mathrm{s}^{2} .\) Find the magnitude and direction of the puck's velocity at a time of \(t=0.50\) s. Specify the direction relative to the \(+x\) axis.

A skateboarder shoots off a ramp with a velocity of \(6.6 \mathrm{m} / \mathrm{s}\), directed at an angle of \(58^{\circ}\) above the horizontal. The end of the ramp is \(1.2 \mathrm{m}\) above the ground. Let the \(x\) axis be parallel to the ground, the \(+y\) direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

A space vehicle is coasting at a constant velocity of \(21.0 \mathrm{m} / \mathrm{s}\) in the \(+y\) direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at \(0.320 \mathrm{m} / \mathrm{s}^{2}\) in the \(+x\) direction. After \(45.0 \mathrm{s},\) the pilot shuts off the \(\mathrm{RCS}\) thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle measured from the \(+y\) direction.

A ferryboat is traveling in a direction \(38.0^{\circ}\) north of east with a speed of \(5.50 \mathrm{m} / \mathrm{s}\) relative to the water. A passenger is walking with a velocity of \(2.50 \mathrm{m} / \mathrm{s}\) due east relative to the boat. What is the velocity (magnitude and direction) of the passenger with respect to the water? Determine the directional angle relative to due east.

Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of \(1.9 \mathrm{m} / \mathrm{s}\) parallel to the ground. Upon contact with the bat the ball is \(1.2 \mathrm{m}\) above the ground. Player \(\mathrm{B}\) wishes to duplicate this bunt, in so far as he also wants to give the ball a velocity parallel to the ground and have his ball travel the same horizontal distance as player A's ball does. However, player B hits the ball when it is \(1.5 \mathrm{m}\) above the ground. What is the magnitude of the initial velocity that player B's ball must be given?
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