/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 \(\bullet\) A player kicks a foo... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\bullet\) A player kicks a football at an angle of \(40.0^{\circ}\) from the horizontal, with an initial speed of 12.0 \(\mathrm{m} / \mathrm{s} .\) A second player standing at a distance of 30.0 \(\mathrm{m}\) from the first (in the direction of the kick) starts running to meet the ball at the instant it is kicked. How fast must he run in order to catch the ball just before it hits the ground?

Short Answer

Expert verified
The player must run at a speed of 9.88 m/s.

Step by step solution

01

Analyze the Problem

We have a projectile motion problem where a ball is kicked at a certain angle. Simultaneously, another person starts running to intercept the ball. We need to determine how fast the second player must run to catch the ball before it lands.
02

Break Down Initial Velocity

The initial velocity of the ball is given as 12.0 m/s, at a 40.0° angle. We need to find the horizontal and vertical components of this initial velocity. Use the formulas:\[ v_{0x} = v_0 \cos(\theta) \] \[ v_{0y} = v_0 \sin(\theta) \] Substitute the given values:\( v_{0x} = 12.0 \cos(40.0^{\circ}) \approx 9.19 \text{ m/s} \) \( v_{0y} = 12.0 \sin(40.0^{\circ}) \approx 7.71 \text{ m/s} \)
03

Calculate Time of Flight

The time the ball spends in the air (time of flight) can be found using the vertical component and the gravity formula for time, where the total vertical displacement is zero when it lands:\[ t = \frac{2v_{0y}}{g} \] Taking \( g = 9.8 \text{ m/s}^2 \):\( t = \frac{2 \times 7.71}{9.8} \approx 1.57 \text{ s} \)
04

Determine Horizontal Range

The horizontal range is the distance the ball travels in the timeframe it is in the air. Use the formula:\[ R = v_{0x} \times t \]Substitute the known values:\( R = 9.19 \times 1.57 \approx 14.42 \text{ m} \)
05

Calculate Runner's Speed

To calculate the speed at which the second player must run, determine the distance he needs to cover (30 m minus the horizontal range) and divide it by the time of flight:\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{30.0 - 14.42}{1.57} \approx 9.88 \text{ m/s} \]

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. In the case of projectile motion, such as a kicked football, we analyze how the object moves through space. We break down its motion into horizontal and vertical components since they are independent of one another. This helps in simplifying calculations.
  • The horizontal motion of a projectile is at a constant speed because there are no horizontal forces acting upon it (assuming air resistance is negligible).
  • The vertical motion, however, is affected by gravity, resulting in an upward deceleration and downward acceleration.
By understanding kinematics, we can predict the future position of an object at any given point in time. For a projectile, this involves determining how far and high it will travel before returning to the ground.
Initial Velocity Calculation
In any projectile problem, identifying the initial velocity's components is critical. The initial velocity is usually given as a single value with a direction, often described by an angle from the horizontal.
To find the components:
  • The horizontal component is found using the equation: \( v_{0x} = v_0 \cos(\theta) \), where \( v_0 \) is the initial speed and \( \theta \) is the launch angle.
  • The vertical component is determined by the equation: \( v_{0y} = v_0 \sin(\theta) \).
For instance, with an initial speed of 12.0 m/s at a 40° angle, the horizontal velocity is approximately 9.19 m/s, and the vertical velocity is approximately 7.71 m/s. These components help proceed with further analysis, including predicting how long the projectile will be in the air.
Time of Flight
The time of flight in projectile motion tells us how long the projectile remains in the air. This is a crucial piece of data when assessing any motion.To determine it, consider the vertical component of motion, because the projectile must return to its original vertical position if shot and landing at the same height.
  • The time of flight formula is derived from: \( t = \frac{2v_{0y}}{g} \), where \( g \) is the acceleration due to gravity (9.8 m/s²).
Using our example, the vertical component \( v_{0y} \) is 7.71 m/s, which results in a time of flight of approximately 1.57 seconds. This duration allows us to determine other factors of the flight, such as how far the projectile travels horizontally.
Horizontal Range
The horizontal range is the total horizontal distance traveled by the projectile during its flight. Calculating this range allows us to understand how far the projectile can land from its starting point.The calculation builds on the horizontal component of the initial velocity and the time of flight.
  • The formula is: \( R = v_{0x} \times t \), where \( v_{0x} \) is the horizontal component found in initial velocity calculations.
In our scenario, with \( v_{0x} \approx 9.19 \text{ m/s} \) and \( t \approx 1.57 \text{ s} \), the horizontal range is approximately 14.42 meters. Knowing the horizontal range is vital for resolving further parts of a problem, such as determining if someone moving from the initial point can intercept the projectile or how much adjustment is needed in the launch angle or speed for achieving a desired distance.

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Most popular questions from this chapter

A physics book slides off a horizontal tabletop with a speed of 1.10 \(\mathrm{m} / \mathrm{s}\) . It strikes the floor in 0.350 s. Ignore air resistance. Find (a) the height of the tabletop above the floor, (b) the horizontal distance from the edge of the table to the point where the book strikes the floor, and (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.

If the time of flight of the ball is \(t\) seconds, at what point in time will the ball have zero vertical velocity? A. \(t / 4 \mathrm{s}\) B. \(t / 4 \mathrm{s}\) C. \(t / 2 \mathrm{s}\) D. \(t / 2 \mathrm{s}\) E. There is no place on the path where the vertical velocity is zero.

A batted baseball leaves the bat at an angle of \(30.0^{\circ}\) above the horizontal and is caught by an outfielder 375 ft from home plate at the same height from which it left the bat. (a) What was the initial speed of the ball? (b) How high does the ball rise above the point where it struck the bat?

\(\cdot\) A model of a helicopter rotor has four blades, each 3.40 \(\mathrm{m}\) in length from the central shaft to the tip of the blade. The model is rotated in a wind tunnel at 550 rev/min. (a) What is the linear speed, in \(\mathrm{m} / \mathrm{s}\) , of the blade tip? (b) What is the radial acceleration of the blade tip, expressed as a multiple of the acceleration \(g\) due to gravity?

\bullet A cart carrying a vertical missile launcher moves horizontally at a constant velocity of 30.0 \(\mathrm{m} / \mathrm{s}\) to the right. It launches a rocket vertically upward. The missile has an initial vertical velocity of 40.0 \(\mathrm{m} / \mathrm{s}\) relative to the cart. (a) How high does the rocket go? (b) How far does the cart travel while the rocket is in the air? (c) Where does the rocket land relative to the cart?

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