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You're standing outside on a windless day when raindrops begin to fall straight down. You run for shelter at a speed of \(5.0 \mathrm{~m} / \mathrm{s},\) and you notice while you're running that the raindrops appear to be falling at an angle of about \(30^{\circ}\) from the vertical. What's the vertical speed of the raindrops?

Short Answer

Expert verified
The vertical speed of the raindrops is approximately 8.66 m/s.

Step by step solution

01

Understand the Problem

You need to find the vertical speed of the raindrops as it appears to be falling at an angle to someone running. Given your running speed and the angle of the apparent fall of the raindrops, you can determine the true vertical speed of the raindrops.
02

Analyze Motion

As you run for shelter, you create a relative motion between you and the rain. This motion causes the rain to appear to be coming at an angle from the vertical. Your running creates a horizontal component of the velocity of the raindrops relative to you.
03

Set Up Vector Components

There are two components in this scenario: horizontal and vertical. The apparent angle of the rain provides the necessary data to decompose the velocity vector of the rain into these components.
04

Formulate the Relative Motion Equations

The apparent angle of the rain, θ = 30°, gives you: \( \tan(30°) = \frac{v_{horizontal}}{v_{vertical}} \). The horizontal component is your running speed, \(v_{horizontal} = 5.0 \mathrm{~m/s} \). Use these to solve for \(v_{vertical}\).
05

Apply Trigonometric Relation

Substitute the known values into the tangent equation: \( \tan(30°) = \frac{5.0}{v_{vertical}} \). Use \( \tan(30°) = \frac{1}{\sqrt{3}} \) to solve for the actual velocity.
06

Calculate the Vertical Speed

Rearrange the equation \( \frac{1}{\sqrt{3}} = \frac{5.0}{v_{vertical}} \) to find \(v_{vertical} = 5.0 \times \sqrt{3} \mathrm{~m/s} \). Solving this gives \(v_{vertical} \approx 8.66 \mathrm{~m/s}\).

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Key Concepts

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

Relative Motion
When you're on the move, the movement of other objects around you can seem different from reality. This is known as relative motion. In our exercise, standing still on a windless day, raindrops fall straight down. But, when you start running to find shelter, the raindrops don't appear to be falling straight anymore. They seem to come at an angle to you.
This change in appearance happens because of your movement. To understand this better, imagine you and the raindrops as two moving objects. Your running creates a distinct velocity, and this affects how the raindrops' motion is perceived.
This concept is crucial because it tells us why the rain seems to come at a 30-degree angle while you're actually moving. Understanding relative motion helps you grasp why movements change depending on the observer's frame of reference.
Trigonometric Relations
Trigonometry often comes up in physics problems involving angles and vectors. In this problem, the focus is on tan (tangent), a basic trigonometric function. The raindrops appear to fall at a 30-degree angle from the vertical, and this provides crucial information to solve the problem.

By using the tangent of an angle, which is defined as the ratio of the opposite side to the adjacent side in a right triangle, we can set up an equation. The angle in our scenario is 30 degrees, so we write:
  • \( \tan(30^\circ) = \frac{v_{horizontal}}{v_{vertical}} \)

Here, the horizontal component corresponds to your running speed (5.0 m/s) and the vertical speed is what we need to find. The value of \( \tan(30^\circ) \) is \( \frac{1}{\sqrt{3}} \), a handy trig fact that can expedite your calculations. Applying these relationships is key to solving for the rain's actual vertical speed.
Velocity Components
Understanding velocities in vectors allows us to decompose them into components. In physics, velocities often break into horizontal and vertical parts. This decomposition helps in isolating the specific motion details.
In this exercise, the raindrops have an actual vertical speed, yet they appear to have a horizontal motion relative to you. As you're running, your speed becomes the horizontal velocity component of the rain's motion.
By setting up the problem with:
  • Horizontal velocity component: 5.0 m/s
  • Angle of motion: 30 degrees
...we form equations to calculate the true vertical velocity. Using the trigonometric relation of tangent, we solve:
  • \( \tan(30^\circ) = \frac{5.0}{v_{vertical}} \)
  • Solving this gives \( v_{vertical} = 5.0 \times \sqrt{3} \) m/s, approximately 8.66 m/s.

The velocity components break down complex motions into simpler parts, which can then be individually analyzed and solved.

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Most popular questions from this chapter

A tennis player hits a ball at ground level, giving it an initial velocity of \(24 \mathrm{~m} / \mathrm{s}\) at \(57^{\circ}\) above the horizontal. (a) What are the horizontal and vertical components of the ball's initial velocity? (b) How high above the ground does the ball go? (c) How long does it take the ball to reach its maximum height? (d) What are the ball's velocity and acceleration at its highest point? (e) For how long a time is the ball in the air? (f) When this ball lands on the court, how far is it from the place where it was hit?

A football is kicked from ground level at a speed of \(25 \mathrm{~m} / \mathrm{s}\). If it reaches a maximum height of \(24 \mathrm{~m},\) at what angle was it kicked (relative to horizontal)?

A football is thrown with an initial upward velocity component of \(15.0 \mathrm{~m} / \mathrm{s}\) and a horizontal velocity component of \(18.0 \mathrm{~m} / \mathrm{s}\). (a) How much time is required for the football to reach the highest point in its trajectory? (b) How high does it get above its release point? (c) How much time after it is thrown does it take to return to its original height? How does this time compare with what you calculated in part (a)? Is your answer reasonable? (d) How far has the football traveled horizontally from its original position?

According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor league game. The ball traveled 188 \(\mathrm{m}(618 \mathrm{ft})\) before landing on the ground outside the ballpark. (a) Assuming that the ball's initial velocity was \(45^{\circ}\) above the horizontal, and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point \(0.9 \mathrm{~m}(3.0 \mathrm{ft})\) above ground level? Assume that the ground was perfectly flat. (b) How far would the ball be above a fence \(3.0 \mathrm{~m}(10 \mathrm{ft})\) in height if the fence were \(116 \mathrm{~m}(380 \mathrm{ft})\) from home plate?

A daring swimmer dives off a cliff with a running horizontal leap, as shown in Figure \(3.35 .\) What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is \(1.75 \mathrm{~m}\) wide and \(9.00 \mathrm{~m}\) below the top of the cliff?

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