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Chapter 4: Problem 14

An astronaut on a strange planet finds that she can jump a maximum horizontal distance of \(15.0 \mathrm{m}\) if her initial speed is \(3.00 \mathrm{m} / \mathrm{s} .\) What is the free-fall acceleration on the planet?

Short Answer

Expert verified
The free-fall acceleration \(g\) on the unknown planet is \(0.60 \, \mathrm{m/s}^2\).

Step by step solution

01

Identify known and unknown values

Firstly, identify the known and unknown values in this problem. Known values are the horizontal distance \(d = 15.0 \mathrm{m}\) and the initial speed \(v_i = 3.00 \, \mathrm{m/s}\). The unknown is the free-fall acceleration \(g\), which we need to find.
02

Select relevant formula

Next, select the formula that can relate these values together. The best formula to achieve this task is the range formula of a projectile motion, \(d = \frac{v_i^2}{g}\), where \(d\) is the maximum horizontal distance, \(v_i\) is the initial speed and \(g\) is the gravity acceleration.
03

Rearrange the formula

We then rearrange the formula to solve for the unknown value \(g\), which gives us \(g = \frac{v_i^2}{d}\).
04

Substitute known values into the formula

Next, we substitute the known values \(d = 15 \, \mathrm{m}\) and \(v_i = 3.00\, m/s\) into the formula, giving \(g = \frac{ (3.00 \, \mathrm{m/s})^2}{15 \, \mathrm{m}}\).
05

Perform the calculation

Quickly calculate the result to get the value of \(g\) on this unknown planet. So, the result of \(g = \frac{(3.00 \, \mathrm{m/s})^2}{15 \, \mathrm{m}} = 0.60 \, \mathrm{m/s^2}\).

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Key Concepts

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

Understanding Projectile Motion
Projectile motion is a type of motion experienced by an object that is thrown or projected into the air. The motion follows a curved trajectory under the influence of gravity. It can be broken down into two components: horizontal and vertical motion.
  • Horizontal Component: The horizontal motion occurs at a constant speed because there is no acceleration (ignoring air resistance). This component does not affect the vertical motion.
  • Vertical Component: The vertical motion is influenced by gravity, which causes an object to accelerate downwards at a constant rate.
By analyzing both components, we can predict the path of the projectile. In our astronaut's scenario, we focus on how far she can jump horizontally, which is part of her projectile motion on this strange planet.
Exploring the Range Formula
The range formula for projectile motion helps determine how far a projectile will travel horizontally. This formula is especially useful when the launch and landing heights are the same.

The formula is given by: \[ d = \frac{v_i^2}{g} \]where:
  • \(d\) is the horizontal distance or range
  • \(v_i\) is the initial speed of the projectile
  • \(g\) is the acceleration due to gravity
In our problem, this formula is used to connect the horizontal distance the astronaut can jump with her initial speed and the planet's gravitational force. Understanding this formula allows for calculating one of the three parameters if the other two are known.
Calculation of Gravity on the Planet
Gravity calculation on a planet can be determined using the range formula. By rearranging the formula to solve for the gravitational acceleration \(g\), we can find the force of gravity on a planet.

Rearranging the formula:\[ g = \frac{v_i^2}{d} \]We substitute the given values: the initial speed of \(3.00 \ \mathrm{m/s}\) and the maximum horizontal distance of \(15.0 \ \mathrm{m}\).

Calculating this gives us: \[ g = \frac{(3.00 \ \mathrm{m/s})^2}{15.0 \ \mathrm{m}} = 0.60 \ \mathrm{m/s^2} \]This calculation reveals that the gravity on this planet is much less than Earth's gravity of \(9.81 \ \mathrm{m/s^2}\), which explains why the astronaut can jump farther.
The Role of Initial Speed
Initial speed is crucial in projectile motion as it affects both how high and how far the projectile will travel. It is the speed with which the projectile is launched.

In our example, the astronaut's initial speed is \(3.00\ \mathrm{m/s}\). This speed determines the maximum range she can achieve when she jumps. The greater the initial speed, the farther the projectile can travel horizontally. This is because the initial speed is squared in the range formula, showing its strong impact on the distance.

Practically, increasing the initial speed could result in a greater distance, as observed in various sports such as javelin throw or basketball, where players aim to increase their initial speed to achieve longer distances.

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Most popular questions from this chapter

It is not possible to see very small objects, such as viruses, using an ordinary light microscope. An electron microscope can view such objects using an electron beam instead of a light beam. Electron microscopy has proved invaluable for investigations of viruses, cell membranes and subcellular structures, bacterial surfaces, visual receptors, chloroplasts, and the contractile properties of muscles. The "lenses" of an electron microscope consist of electric and magnetic fields that control the electron beam. As an example of the manipulation of an electron beam, consider an electron traveling away from the origin along the \(x\) axis in the \(x y\) plane with initial velocity \(\mathbf{v}_{i}=v_{i}\) i. As it passes through the region \(x=0\) to \(x=d\), the electron experiences acceleration \(\mathbf{a}=a_{x} \mathbf{i}+a_{y} \mathbf{j},\) where \(a_{x}\) and \(a_{y}\) are constants. For the case \(v_{i}=1.80 \times 10^{7} \mathrm{m} / \mathrm{s}\) \(a_{x}=8.00 \times 10^{14} \mathrm{m} / \mathrm{s}^{2}\) and \(a_{y}=1.60 \times 10^{15} \mathrm{m} / \mathrm{s}^{2},\) determine at \(x=d=0.0100 \mathrm{m}\) (a) the position of the electron, (b) the velocity of the electron, (c) the speed of the electron, and (d) the direction of travel of the electron (i.e., the angle between its velocity and the \(x\) axis).

A motorist drives south at \(20.0 \mathrm{m} / \mathrm{s}\) for \(3.00 \mathrm{min}\) then turns west and travels at \(25.0 \mathrm{m} / \mathrm{s}\) for \(2.00 \mathrm{min},\) and finally travels northwest at \(30.0 \mathrm{m} / \mathrm{s}\) for 1.00 min. For this 6.00 -min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity. Let the positive \(x\) axis point east.

A Coast Guard cutter detects an unidentified ship at a distance of \(20.0 \mathrm{km}\) in the direction \(15.0^{\circ}\) east of north. The ship is traveling at \(26.0 \mathrm{km} / \mathrm{h}\) on a course at \(40.0^{\circ}\) east of north. The Coast Guard wishes to send a speedboat to intercept the vessel and investigate it. If the speedboat travels \(50.0 \mathrm{km} / \mathrm{h},\) in what direction should it head? Express the direction as a compass bearing with respect to due north.

Two swimmers, Alan and Beth, start together at the same point on the bank of a wide stream that flows with a speed v. Both move at the same speed \(c(c>v),\) relative to the water. Alan swims downstream a distance \(L\) and then \(\mathrm{up}^{-}\) stream the same distance. Beth swims so that her motion relative to the Earth is perpendicular to the banks of the stream. She swims the distance \(L\) and then back the same distance, so that both swimmers return to the starting point. Which swimmer returns first? (Note: First guess the answer.)

The small archerfish (length 20 to \(25 \mathrm{cm}\) ) lives in brackish waters of southeast Asia from India to the Philippines. This aptly named creature captures its prey by shooting a stream of water drops at an insect, either flying or at rest. The bug falls into the water and the fish gobbles it up. The archerfish has high accuracy at distances of \(1.2 \mathrm{m}\) to \(1.5 \mathrm{m}\) and it sometimes makes hits at distances up to 3.5 m. A groove in the roof of its mouth, along with a curled tongue, forms a tube that enables the fish to impart high velocity to the water in its mouth when it suddenly closes its gill flaps. Suppose the archerfish shoots at a target \(2.00 \mathrm{m}\) away, at an angle of \(30.0^{\circ}\) above the horizontal. With what velocity must the water stream be launched if it is not to drop more than \(3.00 \mathrm{cm}\) vertically on its path to the target?
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