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Chapter 2: Problem 54

A ball is thrown upward from the top of a 25.0 -m-tall building. The ball's initial speed is \(12.0 \mathrm{~m} / \mathrm{s}\). At the same instant, a person is running on the ground at a distance of \(31.0 \mathrm{~m}\) from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

Short Answer

Expert verified
The runner needs an average speed of about 22.96 m/s.

Step by step solution

01

Determine the Time for the Ball to Fall

We start by determining the total time it takes for the ball to hit the ground. The ball is thrown upward with an initial speed of \(12.0\, \text{m/s}\) from a height of \(25.0\, \text{m}\). Using the equation for vertical motion \(h = v_i t + \frac{1}{2} a t^2\), we have:\[0 = 25.0 + 12.0t - \frac{1}{2}(9.8)t^2\]Which simplifies to the quadratic equation:\[0 = 12.0t - 4.9t^2 - 25.0\]
02

Solve the Quadratic Equation

To solve \(0 = 12.0t - 4.9t^2 - 25.0\), use the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = -4.9\), \(b = 12.0\), and \(c = -25.0\):\[t = \frac{-12.0 \pm \sqrt{12.0^2 - 4(-4.9)(-25.0)}}{2(-4.9)}\]\[t = \frac{-12.0 \pm \sqrt{144 - 490}}{-9.8}\]\[t = \frac{-12.0 \pm \sqrt{-346}}{-9.8}\]Fix the equation:\[t = \frac{-12.0 \pm \sqrt{144 + 490}}{-9.8}\]\[t = \frac{-12.0 \pm \sqrt{634}}{-9.8}\]Calculate \(t\).
03

Calculate the Time for the Ball to Reach Ground

From the simplified equation \(t = \frac{-12.0 \pm \sqrt{634}}{-9.8}\), we compute the positive time:\[t = \frac{-12.0 + \sqrt{634}}{-9.8}\]Calculating, \(\sqrt{634} \approx 25.2\), thus:\[t = \frac{25.2 - 12.0}{9.8}\]\[t \approx \frac{13.2}{9.8} \approx 1.35\, \text{seconds}\]
04

Calculate the Required Speed of the Runner

The person needs to cover \(31.0\, \text{m}\) in the same time, \(1.35\, \text{seconds}\). The average speed \(v\) is given by \(v = \frac{d}{t}\), where \(d = 31.0\, \text{m}\) and \(t \approx 1.35\, \text{s}\):\[v = \frac{31.0}{1.35} \approx 22.96\, \text{m/s}\]
05

Conclusion

To catch the ball at the base of the building, the runner must maintain an average speed of approximately \(22.96\, \text{m/s}\).

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Key Concepts

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

Quadratic Equations in Physics
In physics, quadratic equations often appear when dealing with motion under constant acceleration. A typical scenario involves projectile motion, like a ball being thrown into the air. When analyzing such problems, we use the kinematic equation:
  • \( h = v_i t + \frac{1}{2} a t^2 \)
Here, \( h \) is the height, \( v_i \) is the initial velocity, \( a \) is the acceleration due to gravity, and \( t \) represents time. Given these parameters form a quadratic equation when rearranged for time, solutions are found using the quadratic formula:
  • \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
When solving, it's essential to pick the root that makes sense physically, often the positive time value. Quadratics can look complex in physics, but breaking down each term and ensuring all units align can simplify the process.
Kinematic Equations
Kinematic equations describe the motion of objects and relate displacement, velocity, acceleration, and time. The key equations include:
  • \( v = v_i + at \)
  • \( s = v_i t + \frac{1}{2} a t^2 \)
  • \( v^2 = v_i^2 + 2as \)
In these, \( v \) represents final velocity, \( v_i \) is initial velocity, \( a \) is acceleration, \( t \) is time, and \( s \) is displacement. Projectile motion, a common use case, involves both horizontal and vertical components. The vertical equation helps find how long an object stays in flight when gravity is the only acting force. This involves using a quadratic equation to solve for time, as seen in the step-by-step solution. Understanding these equations and their interrelationships is crucial for accurately analyzing movement.
Average Speed Calculation
Calculating average speed is an essential part of understanding motion. It involves dividing the total distance traveled by the total time taken to cover that distance:
  • \( v = \frac{d}{t} \)
Where \( v \) is average speed, \( d \) is the distance, and \( t \) is time. In the context of our problem, this calculation helps determine how fast a person must run to meet the ball at a specific point, ensuring the average speed is maintained over the entire distance.
When calculating, it's important to ensure time is consistent with the context of the problem, such as how long the ball is in motion. Problems like these illustrate the direct application of average speed in physics to predict outcomes in dynamic situations.

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Most popular questions from this chapter

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The stones are thrown with the same speed. (a) Does the stone thrown upward gain or lose speed as it moves upward? Why? (b) Does the stone thrown downward gain or lose speed as time passes? Explain. (c) The speed at which the stones are thrown is such that they cross paths. Where do they cross paths, above, at, or below the point that corresponds to half the height of the cliff? Justify your answer. Problem The height of the cliff is \(6.00 \mathrm{~m},\) and the speed with which the stones are thrown is \(9.00 \mathrm{~m} / \mathrm{s}\). Find the location of the crossing point. Check to see that your answer is consistent with your answers to the Concept Questions.

In 1954 the English runner Roger Bannister broke the four-minute barrier for the mile with a time of 3: 59.4 s \((3\) min and \(59.4 \mathrm{~s}\) ). In 1999 the Moroccan runner Hicham elGuerrouj set a record of 3: 43.13 s for the mile. If these two runners had run in the same race, each running the entire race at the average speed that earned him a place in the record books, el-Guerrouj would have won. By how many meters?

In preparation for this problem, review Conceptual Example 7 . From the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The initial speed of the pebble is \(9.0 \mathrm{~m} / \mathrm{s}\). (a) What is the acceleration (magnitude and direction) of the pebble during the downward motion? Is the pebble decelerating? Explain. (b) After \(0.50 \mathrm{~s}\), how far beneath the cliff top is the pebble?

Two cars cover the same distance in a straight line. Car \(\mathrm{A}\) covers the distance at a constant velocity. Car B starts from rest, maintains a constant acceleration, and covers the distance in the same time as car A. (a) Is car B's final velocity smaller than, equal to, or greater than car A's constant velocity? Explain. (b) Is car B's average velocity smaller than, equal to, or greater than car A's constant velocity? Why? (c) Given a value for the time to cover the distance and the fact that car B starts from rest, what other variable must be known before car B's acceleration can be determined? Provide a reason for your answer. Problem Both cars cover a distance of \(460 \mathrm{~m}\) in \(210 \mathrm{~s}\). Assume that they are moving in the \(+x\) direction. Determine (a) the constant velocity of car \(\mathrm{A},\) (b) the final velocity of car \(\mathrm{B},\) and \((\mathrm{c})\) the acceleration of car \(\mathrm{B}\). Verify that your answers are consistent with your answers to the Concept Questions.

A car makes a trip due north for three-fourths of the time and due south one- fourth of the time. The average northward velocity has a magnitude of \(27 \mathrm{~m} / \mathrm{s},\) and the average southward velocity has a magnitude of \(17 \mathrm{~m} / \mathrm{s}\). What is the average velocity, magnitude and direction, for the entire trip?
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