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Chapter 4: Problem 30

A soccer ball is kicked from the ground with an initial speed of \(19.5 \mathrm{~m} / \mathrm{s}\) at an upward angle of \(45^{\circ} .\) A player \(55 \mathrm{~m}\) away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground?

Short Answer

Expert verified
The player must run at an average speed of approximately 5.79 m/s.

Step by step solution

01

Find the Time of Flight

First, decompose the initial velocity into horizontal and vertical components using the angle 45掳.The horizontal and vertical components are given by:\[ v_{x0} = v_0 \cos(45^{\circ}) = 19.5 \times \frac{\sqrt{2}}{2} \approx 13.79 \text{ m/s}\]\[ v_{y0} = v_0 \sin(45^{\circ}) = 19.5 \times \frac{\sqrt{2}}{2} \approx 13.79 \text{ m/s}\]Using the vertical motion equation, find the time of flight:\[ t_{total} = \frac{2v_{y0}}{g} = \frac{2 \times 13.79}{9.8} \approx 2.81 \text{ seconds}\]
02

Calculate Horizontal Distance Traveled by the Ball

To find out how far the ball travels horizontally, use the horizontal component of the velocity:\[ d_{ball} = v_{x0} \times t_{total} = 13.79 \times 2.81 \approx 38.74 \text{ meters} \]
03

Compute the Player's Required Speed

Since the ball travels only 38.74 meters in the air, the player, initially 55 meters away, needs to run a shorter distance.Subtract the ball's travel distance from the player's starting distance:\[ d_{player} = 55 - 38.74 = 16.26 \text{ meters} \]Calculate the average speed required for the player to run this distance:\[ v_{player} = \frac{d_{player}}{t_{total}} = \frac{16.26}{2.81} \approx 5.79 \text{ m/s} \]

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Key Concepts

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

Initial Velocity Components
In projectile motion, the initial velocity is always split into two components: horizontal ( \(v_{x0}\) ) and vertical ( \(v_{y0}\) ). For a soccer ball kicked with an initial speed, understanding these components is key to predicting its path. These components are calculated using trigonometric functions on the angle of projection, which is 45掳 in this case.
The horizontal component ( \(v_{x0}\)) is determined by:
  • \(v_{x0} = v_0 \cos(\theta)\)
  • Where \(v_0\) is the initial speed and \(\theta\) is the angle of elevation.
  • With \(v_0 = 19.5 \, \text{m/s}\) and \(\theta = 45掳\), it evaluates to \(13.79 \, \text{m/s}\).
The vertical component ( \(v_{y0}\) ) is:
  • \(v_{y0} = v_0 \sin(\theta)\)
  • This also results in 13.79 m/s, given the symmetrical nature of the 45掳 angle.
The symmetry in these components greatly simplifies calculations, as seen in this exercise.
It allows us to see clearly how both the horizontal and vertical motions contribute to the overall motion of the ball.
Time of Flight
The time of flight for a projectile is crucial to determining how long it stays in the air. It is entirely dependent on the vertical motion of the projectile. In this scenario, the time it takes for the ball to rise and then fall back to the ground is calculated using the vertical velocity component and gravity.
The formula used is:
  • \(t_{total} = \frac{2v_{y0}}{g}\)
  • Where \(v_{y0}\) is the initial vertical velocity and \(g\) is the acceleration due to gravity (9.8 m/s虏).
  • This calculation yields a total time of flight for the soccer ball of approximately 2.81 seconds.
Breaking this down helps illustrate that with a 45掳 launch angle and equal horizontal and vertical components, the path and time the ball stays airborne becomes predictable.
This constancy holds true for any initial speed and launch angle yielding such symmetric components, provided air resistance is negligible.
Horizontal and Vertical Motion
In the realm of projectile motion, understanding horizontal and vertical movements separately can simplify analysis.
The horizontal motion:
  • It's driven by the horizontal velocity component, \(v_{x0}\), which in this scenario, remains constant since there are no forces like air resistance in this idealized environment.
  • The distance traveled horizontally is given by \(d_{ball} = v_{x0} \times t_{total}\) .
  • In this exercise, the ball travels approximately 38.74 meters horizontally.
The vertical motion:
  • It's affected solely by gravity, which acts to decelerate the ball on the way up and accelerate it on the way down.
  • The peak and subsequent return to the ground all occur over the calculated time of flight, 2.81 seconds.
Grasping this distinct separation of the two motions allows students to analyze each independently.
This demystifies the trajectory and ensures more accurate predictions of where and when the projectile will hit the ground.
Visualizing these separate paths and forces acting on the ball enhances a student's ability to solve related physics problems.

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Most popular questions from this chapter

Long flights at midlatitudes in the Northern Hemisphere encounter the jet stream, an eastward airflow that can affect a plane's speed relative to Earth's surface. If a pilot maintains a certain speed relative to the air (the plane's airspeed), the speed relative to the surface (the plane's ground speed) is more when the flight is in the direction of the jet stream and less when the flight is opposite the jet stream. Suppose a round-trip flight is scheduled between two cities separated by \(4000 \mathrm{~km}\), with the outgoing flight in the direction of the jet stream and the return flight opposite it. The airline computer advises an airspeed of \(1000 \mathrm{~km} / \mathrm{h},\) for which the difference in flight times for the outgoing and return flights is \(70.0 \mathrm{~min} .\) What jet-stream speed is the computer using?

A boat is traveling upstream in the positive direction of an \(x\) axis at \(14 \mathrm{~km} / \mathrm{h}\) with respect to the water of a river. The water is flowing at \(9.0 \mathrm{~km} / \mathrm{h}\) with respect to the ground. What are the (a) magnitude and (b) direction of the boat's velocity with respect to the ground? A child on the boat walks from front to rear at \(6.0 \mathrm{~km} / \mathrm{h}\) with respect to the boat. What are the (c) magnitude and (d) direction of the child's velocity with respect to the ground?

A cannon located at sea level fires a ball with initial speed \(82 \mathrm{~m} / \mathrm{s}\) and initial angle \(45^{\circ} .\) The ball lands in the water after traveling a horizontal distance \(686 \mathrm{~m} .\) How much greater would the horizontal distance have been had the cannon been \(30 \mathrm{~m}\) higher?

A magnetic field forces an electron to move in a circle with radial acceleration \(3.0 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} .\) (a) What is the speed of the electron if the radius of its circular path is \(15 \mathrm{~cm} ?\) (b) What is the period of the motion?

A cat rides a merry-go-round turning with uniform circular motion. At time \(t_{1}=2.00 \mathrm{~s},\) the cat's velocity is \(\vec{v}_{1}=\) \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}},\) measured on a horizontal \(x y\) coordinate system. At \(t_{2}=5.00 \mathrm{~s},\) the cat's velocity is \(\vec{v}_{2}=(-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+\) \((-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) What are (a) the magnitude of the cat's centripetal acceleration and (b) the cat's average acceleration during the time interval \(t_{2}-t_{1},\) which is less than one period?
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