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Chapter 4: Problem 30

A soccer ball is kicked from the ground with an initial speed of \(21.3 \mathrm{~m} / \mathrm{s}\) 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 2.75 m/s.

Step by step solution

01

Determine Time of Flight

First, we need to calculate the time the soccer ball is in the air. We'll use the formula for the time of flight for a projectile: \[ t = \frac{2v_{0} \sin(\theta)}{g} \]where \(v_{0} = 21.3\, m/s\), \(\theta = 45^\circ\), and \(g = 9.8\, m/s^2\). Thus, \[ t = \frac{2 \times 21.3 \times \sin(45^\circ)}{9.8} = \frac{30.09}{9.8} \approx 3.07 \text{ s} \]
02

Calculate Horizontal Distance Covered by the Ball

The horizontal distance \(R\) covered by the ball is given by \[ R = v_{0} \cos(\theta) \times t \]Substitute the known values:\[ R = 21.3 \times \cos(45^\circ) \times 3.07 \approx 46.56 \text{ m} \]
03

Determine Distance Player Must Cover

The player is 55 meters away in the direction of the kick. To meet the ball, he needs to cover the distance:\[ 55 - 46.56 = 8.44 \text{ m} \]
04

Calculate Player's Required Average Speed

The player needs to cover 8.44 meters in the time the ball is in the air, which is 3.07 seconds. Therefore, his average speed \(v_{p}\) is:\[ v_{p} = \frac{8.44}{3.07} \approx 2.75 \text{ m/s} \]

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Key Concepts

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

Time of Flight
Projectile motion is an intriguing topic in physics, and one of its key aspects is understanding the 'Time of Flight'. The Time of Flight of a projectile is the total time the object spent traveling after it has been launched into the air until it touches the ground again.

In this exercise, a soccer ball is kicked with an initial speed of 21.3 m/s at a 45-degree angle. To calculate the Time of Flight, we use the formula: \[ t = \frac{2v_{0} \sin(\theta)}{g} \]Here, \( v_{0} \) stands for initial velocity, \( \theta \) represents the launch angle, and \( g \) denotes the acceleration due to gravity which is 9.8 m/s虏.

This formula helps us understand how long the ball will remain airborne by considering the vertical component of the launch velocity. By substituting the values into the equation, we find that the ball stays in the air for approximately 3.07 seconds. This comprehension allows us to predict the flight duration of any similar projectile under identical conditions.
Horizontal Distance
Another key concept of projectile motion is the calculation of Horizontal Distance or range. This is the distance an object travels along the horizontal plane after being projected into the air.

The formula for Horizontal Distance is:\[ R = v_{0} \cos(\theta) \times t \]In this scenario, we use the initial speed, the angle of projection, and the Time of Flight. \( \cos(\theta) \) helps us compute the portion of the initial speed applied in the horizontal direction.
  • The initial velocity \( v_{0} \) is 21.3 m/s.
  • The angle \( \theta \) is 45 degrees.
  • The Time of Flight is approximately 3.07 seconds.
Substituting these values, we find that the Horizontal Distance covered by the soccer ball is around 46.56 meters.

This insight into the range enables us to determine the landing spot of projectiles, such as predicting where a launched soccer ball might hit the ground.
Average Speed
Average Speed is another vital concept, particularly when it comes to someone or something trying to intercept a projectile. In this exercise, a player needs to cover a certain distance to meet the soccer ball before it lands.

The player, standing 55 meters from the kick-off point, must travel the difference between his starting position and where the ball lands, which we calculated as approximately 46.56 meters away.

Thus, the distance he needs to cover is:\[ 55 - 46.56 = 8.44 \text{ meters} \]To find out how fast he should run to reach the ball in time, we use the concept of Average Speed. The formula for Average Speed over a distance is:\[ v_{p} = \frac{\text{distance}}{\text{time}} \]Substituting our values gives:
  • Distance: 8.44 meters.
  • Time: 3.07 seconds, as calculated from the Time of Flight.
Hence, the player's required Average Speed is approximately 2.75 m/s.

Understanding Average Speed in relation to projectile motion helps analyze scenarios where timing and synchronization are crucial, such as catching a ball or evading obstacles in motion.

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Most popular questions from this chapter

In basketball, hang is an illusion in which a player seems to weaken the gravitational acceleration while in midair. The illusion depends much on a skilled player's ability to rapidly shift the ball between hands during the flight, but it might also be supported by the longer horizontal distance the player travels in the upper part of the jump than in the lower part. If a player jumps with an initial speed of \(v_{0}=6.00 \mathrm{~m} / \mathrm{s}\) at an angle of \(\theta_{0}=35.0^{\circ}\), what percentage of the jump's range does the player spend in the upper half of the jump (between maximum height and half maximum height)?

When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about \(20 \mathrm{~km}\) (about the size of the San Francisco area). If a neutron star rotates once every second, (a) what is the speed of a particle on the star's equator and (b) what is the magnitude of the particle's centripetal acceleration? (c) If the neutron star rotates faster, do the answers to (a) and (b) increase, decrease, or remain the same?

A baseball leaves a pitcher's hand horizontally at a speed of \(153 \mathrm{~km} / \mathrm{h}\). The distance to the batter is \(18.3 \mathrm{~m}\). (a) How long does the ball take to travel the first half of that distance? (b) The second half? (c) How far does the ball fall freely during the first half? (d) During the second half? (e) Why aren't the quantities in \((\mathrm{c})\) and \((\mathrm{d})\) equal?

A watermelon seed has the following coordinates: \(x=-5.0 \mathrm{~m}\), \(y=9.0 \mathrm{~m}\), and \(z=0 \mathrm{~m}\). Find its position vector (a) in unit- vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the \(x\) axis. (d) Sketch the vector on a right-handed coordinate system. If the seed is moved to the \(x y z\) coordinates \((3.00 \mathrm{~m}\), \(0 \mathrm{~m}, 0 \mathrm{~m}\) ), what is its displacement (e) in unit-vector notation and as (f) a magnitude and \((\mathrm{g})\) an angle relative to the positive \(x\) direction?

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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