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Chapter 4: Problem 24

In the 1991 World Track and Field Championships in Tokyo, Mike Powell jumped \(8.95 \mathrm{~m}\), breaking by a full \(5 \mathrm{~cm}\) the 23-year long-jump record set by Bob Beamon. Assume that Powell's speed on takeoff was \(9.5 \mathrm{~m} / \mathrm{s}\) (about equal to that of a sprinter) and that \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\) in Tokyo. How much less was Powell's range than the maximum possible range for a particle launched at the same speed?

Short Answer

Expert verified
Powell's jump was 0.259 m less than the maximum possible range.

Step by step solution

01

Convert Conditions

Given that Powell's speed at takeoff is 9.5 m/s. The record-break distance is 8.95 m, and the acceleration due to gravity \(g\) is 9.8 m/s虏.
02

Understanding Projectile Motion

The maximum range, \(R_{max}\), of a projectile is achieved when it is launched at an angle of 45 degrees. The formula for the range \(R\) of a projectile is given by:\[R = \frac{v^2 \sin(2\theta)}{g}\]where \(v\) is the launch speed and \(\theta\) is the launch angle.
03

Calculate Maximum Possible Range

When \(\theta = 45^\circ\), \(\sin(2\theta) = 1\), the formula reduces to:\[R_{max} = \frac{v^2}{g}\]Substitute \(v = 9.5 \ m/s\) and \(g = 9.8 \ m/s^2\):\[R_{max} = \frac{(9.5)^2}{9.8} = \frac{90.25}{9.8} \approx 9.209 \ m\]
04

Calculate Difference in Range

Powell's jump was 8.95 m. The potential maximum range for the same speed is calculated as 9.209 m.Calculate the difference:\[\Delta R = R_{max} - R_{actual} = 9.209 \ m - 8.95 \ m = 0.259 \ m\]This shows that Powell's jump is 0.259 m less than the maximum possible range.

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Key Concepts

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

Range of Projectile
The range of a projectile is the horizontal distance that a projectile covers during its motion. It depends on multiple factors such as the initial speed, the launch angle, and the acceleration due to gravity. In simpler terms, it's how far the projectile goes horizontally from the point of launch to the point where it lands. To calculate this range, we use the formula:\[R = \frac{v^2 \sin(2\theta)}{g}\]where:
  • \(R\) is the range of the projectile,
  • \(v\) is the initial launch speed,
  • \(\theta\) is the launch angle,
  • \(g\) is the acceleration due to gravity.
This formula indicates that the range is not only dependent on how fast the projectile is launched but also on the angle. Launching at 45 degrees yields the maximum range for a given speed.
Launch Angle
The launch angle, denoted as \(\theta\), is the angle at which a projectile is set into motion with respect to the horizontal. It plays a critical role in determining the maximum range of a projectile. For many practical situations, the launch angle can drastically influence the distance. When the angle is set at 45 degrees, it is known to maximize the range of an ideal projectile, ignoring air resistance. This is due to the mathematical property that \(\sin(2\times 45^\circ)\) equals 1, optimizing the horizontal component of the projectile's velocity effectively.However, anything different from this, either lesser or greater than 45 degrees, will result in a reduced range. Hence, understanding and finding the optimal launch angle is crucial for achieving desired distances, especially in sports or entertainment fields like basketball or track and field.
Acceleration Due to Gravity
The term "acceleration due to gravity" refers to the acceleration that is imparted to objects due to Earth's gravitational pull. It's denoted by \(g\) and typically averages about 9.8 m/s虏 on Earth. In the context of projectile motion, \(g\) is a crucial factor as it influences how quickly a projectile is pulled back toward the earth. In the case of Mike Powell's record-breaking long jump, the value of \(g = 9.8\) m/s虏 was used, which is a pretty standard value for most calculations unless otherwise specified, like at different altitudes or celestial bodies. Understanding \(g\) helps not only to calculate how a projectile falls back down but also helps determine the maximum attainable height and the time it remains in the air. Without the constant of gravity, a projectile would move indefinitely in its motion path, making real-world applications entirely unpredictable.

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Most popular questions from this chapter

Two ships, \(A\) and \(B\), leave port at the same time. Ship \(A\) travels northwest at 24 knots, and ship \(B\) travels at 28 knots in a direction \(40^{\circ}\) west of south \((1\) knot \(=1\) nautical mile per hour; see Appendix D). What are the (a) magnitude and (b) direction of the velocity of ship \(A\) relative to \(B\) ? (c) After what time will the ships be 160 nautical miles apart? (d) What will be the bearing of \(B\) (the direction of \(B\) 's position) relative to \(A\) at that time?

A moderate wind accelerates a pebble over a horizontal \(x y\) plane with a constant acceleration \(\vec{a}=\left(5.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(7.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At time \(t=0\), the velocity is \((4.00 \mathrm{~m} / \mathrm{s}) \hat{i}\). What are the (a) magnitude and (b) angle of its velocity when it has been displaced by \(10.0\) \(\mathrm{m}\) parallel to the \(x\) axis?

A train travels due south at \(30 \mathrm{~m} / \mathrm{s}\) (relative to the ground) in a rain that is blown toward the south by the wind. The path of each raindrop makes an angle of \(70^{\circ}\) with the vertical, as measured by an observer stationary on the ground. An observer on the train, however, sees the drops fall perfectly vertically. Determine the speed of the raindrops relative to the ground.

A proton initially has \(\vec{v}=4.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{j}}+3.0 \hat{\mathrm{k}}\) and then \(4.0 \mathrm{~s}\) later has \(\vec{v}=-2.0 \mathrm{i}-2.0 \hat{\mathrm{j}}+5.0 \hat{\mathrm{k}}\) (in meters per second). For that \(4.0 \mathrm{~s}\), what are (a) the proton's average acceleration \(\vec{a}_{\mathrm{arg}}\) in unitvector notation, (b) the magnitude of \(\vec{a}_{\text {avg }}\), and (c) the angle between \(\vec{a}_{\mathrm{avg}}\) and the positive direction of the \(x\) axis?

A cameraman on a pickup truck is traveling westward at \(20 \mathrm{~km} / \mathrm{h}\) while he records a cheetah that is moving westward \(30 \mathrm{~km} / \mathrm{h}\) faster than the truck. Suddenly, the cheetah stops, turns, and then runs at \(45 \mathrm{~km} / \mathrm{h}\) eastward, as measured by a suddenly nervous crew member who stands alongside the cheetah's path. The change in the animal's velocity takes \(2.0 \mathrm{~s}\). What are the (a) magnitude and (b) direction of the animal's acceleration according to the cameraman and the (c) magnitude and (d) direction according to the nervous crew member?
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