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Chapter 2: Problem 83

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snohomish River below. The swimmers stepped off the bridge, and I estimated that they hit the water 1.5 s later. (a) How high was the bridge? (b) How fast were the swimmers moving when they hit the water? (c) What would the swimmers " drop time be if the bridge were twice as high?

Short Answer

Expert verified
The bridge was about 11 meters high; swimmers hit the water at 14.7 m/s. Double the height has a drop time of ~2.12 s.

Step by step solution

01

Identify Given Information

We know the time of fall is 1.5 seconds, and for calculations on Earth, we generally regard the acceleration due to gravity, \( g \), to be approximately \( 9.8 \, \text{m/s}^2 \).
02

Calculate the Height of the Bridge

Use the formula for the height of a free-falling object without initial velocity: \( h = \frac{1}{2}gt^2 \). Substituting the given values, \( g = 9.8 \, \text{m/s}^2 \) and \( t = 1.5 \, ext{s} \), we find:\[ h = \frac{1}{2} \times 9.8 \times (1.5)^2 = 11.025 \, \text{meters} \]. Thus, the height of the bridge is approximately 11 meters.
03

Calculate the Speed at Impact

The speed of the swimmers when they hit the water can be found with the formula \( v = gt \). Using \( g = 9.8 \, \text{m/s}^2 \) and \( t = 1.5 \, ext{s} \):\[ v = 9.8 \times 1.5 = 14.7 \, \text{m/s} \]. Thus, they were moving at 14.7 meters per second upon impact.
04

Determine Drop Time for Double the Height

For double the height, the new height \( h' = 2h = 2 \times 11 \). Using the second equation for height, \( t' = \sqrt{\frac{2h'}{g}} \): \[ t' = \sqrt{\frac{2 \times 22.05}{9.8}} = \sqrt{4.5} \approx 2.12 \text{ seconds}\]. Thus, the new drop time would be approximately 2.12 seconds.

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Key Concepts

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

Free Fall
Free fall is a specific type of motion in physics where an object is subject to gravity alone, without any other forces such as air resistance. This happens when an object is dropped from a certain height, and it accelerates downwards solely due to the gravitational pull of the Earth. In the original exercise, when swimmers jump off a bridge, they are in free fall as soon as they leave the bridge.

This means their motion is only influenced by gravity, and they experience a constant acceleration as they descend. The principle of free fall is crucial in understanding the swimmers鈥� journey from the bridge to the water. Since true "free fall" assumes no air resistance, we often calculate it under ideal conditions to simplify real-world problems.
Kinematics Equations
Kinematics equations are mathematical formulas used to calculate various aspects of motion such as displacement, velocity, and acceleration. These equations help us to predict how an object will move when it is subjected to certain forces. They are essential in solving problems involving free-falling objects.
  • For example, in our exercise, the height of the bridge was calculated using the equation:

    Height of a Free-Falling Object

    \( h = \frac{1}{2}gt^2 \). This formula tells us how high the bridge is based on how long it took for the swimmers to hit the water and the constant acceleration due to gravity.
  • Meanwhile, the speed of the swimmers upon impact can be determined using another kinematics equation:

    Final Velocity

    \( v = gt \). This helps in understanding how fast an object moves just before it stops falling.
Understanding these equations is key in physics to predict motion accurately.
Acceleration due to Gravity
Acceleration due to gravity is the rate at which an object accelerates when it is in free fall towards the Earth, represented by the symbol \( g \). On Earth, this value is approximately \( 9.8 \, \text{m/s}^2 \). It indicates that with each passing second, the velocity of a free-falling object increases by 9.8 meters per second.

This value is crucial in calculating both the height of the bridge and the speed of impact in our exercise. It is a constant that appears in kinematic equations and gives shape to the motion of falling objects. Acceleration due to gravity allows us to model the free fall of the swimmers accurately along with helping to determine how their motion changes over time. Understanding \( g \) enables us to comprehend why objects fall faster the longer they have been falling.
  • It affects both the time and the velocity calculations for the swimmers as they jump from different heights.

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Most popular questions from this chapter

Two cars drive on a straight highway. At time \(t=0,\) car 1 passes mile marker 0 traveling due east with a speed of \(20.0 \mathrm{m} / \mathrm{s} .\) At the same time, car 2 is \(1.0 \mathrm{km}\) east of mile marker 0 traveling at \(30.0 \mathrm{m} / \mathrm{s}\) due west. Car 1 is speeding up with an acceleration of magnitude \(2.5 \mathrm{m} / \mathrm{s}^{2}\), and car 2 is slowing down with an acceleration of magnitude \(3.2 \mathrm{m} / \mathrm{s}^{2}\) (a) Write \(x\) -versust equations of motion for both cars, taking east as the positive direction. (b) At what time do the cars pass next to one another?

Kangaroos have been clocked at speeds of \(65 \mathrm{km} / \mathrm{h}\). (a) How far can a kangaroo hop in 3.2 minutes at this speed? (b) How long will it take a kangaroo to hop \(0.25 \mathrm{km}\) at this speed?

Two bows shoot identical arrows with the same launch speed. To accomplish this, the string in bow 1 must be pulled back farther when shooting its arrow than the string in bow 2 . (a) Is the acceleration of the arrow shot by bow 1 greater than, less than, or equal to the acceleration of the arrow shot by bow \(2 ?\) (b) Choose the best explanation from among the following: The arrow in bow 2 accelerates for a greater time. II. Both arrows start from rest. III. The arrow in bow 1 accelerates for a greater time.

A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 3.0 s the rocket is at a height of \(77 \mathrm{m}\). (a) What are the magnitude and direction of the rocket's acceleration? (b) What is its speed at this time?

IP You are driving through town at \(12.0 \mathrm{m} / \mathrm{s}\) when suddenly a ball rolls out in front of you. You apply the brakes and begin decelerating at \(3.5 \mathrm{m} / \mathrm{s}^{2}\) (a) How far do you travel before stopping? (b) When you have traveled only half the distance in part \((a),\) is your speed \(6.0 \mathrm{m} / \mathrm{s},\) greater than \(6.0 \mathrm{m} / \mathrm{s},\) or less than \(6.0 \mathrm{m} / \mathrm{s}\) ? Support your answer with a calculation.
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