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Chapter 4: Problem 31

A soccer player claims that he can kick the ball over a wall of height \(3.5 \mathrm{~m}\), which is \(32 \mathrm{~m}\) away along a horizontal field. The player punts the ball from an elevation of \(1.0 \mathrm{~m}\) and the ball is projected at an initial speed of \(18 \mathrm{~m} / \mathrm{s}\) in the direction \(40^{\circ}\) from the horizontal. Does the ball clear the wall?

Short Answer

Expert verified
The ball does not clear the 3.5 m wall.

Step by step solution

01

Breakdown Initial Velocity

Convert the initial velocity into horizontal and vertical components. The initial speed is given by \( v_0 = 18 \mathrm{~m/s} \) and the angle of projection as \( 40^{\circ} \). Use the following equations:\[ v_{0x} = v_0 \cdot \cos(40^{\circ}) \]\[ v_{0y} = v_0 \cdot \sin(40^{\circ}) \]Thus, calculating these gives us:\[ v_{0x} = 18 \cdot \cos(40^{\circ}) \approx 13.79 \mathrm{~m/s} \]\[ v_{0y} = 18 \cdot \sin(40^{\circ}) \approx 11.57 \mathrm{~m/s} \]
02

Find Time to Reach the Wall

Using the horizontal component, we find the time it takes for the ball to reach the wall by using:\[ x = v_{0x} \cdot t_w \]Given \( x = 32 \mathrm{~m} \):\[ t_w = \frac{x}{v_{0x}} = \frac{32}{13.79} \approx 2.32 \mathrm{~s} \]
03

Calculate the Vertical Position

Determine the vertical height of the ball when it reaches the wall. Use the vertical motion equation:\[ y = y_0 + v_{0y} \cdot t_w - \frac{1}{2} g t_w^2 \]where \( y_0 = 1.0 \mathrm{~m} \) and \( g = 9.8 \mathrm{~m/s^2} \) is the acceleration due to gravity:\[ y = 1.0 + 11.57 \cdot 2.32 - \frac{1}{2} \cdot 9.8 \cdot (2.32)^2 \]Calculate this to find the height:\[ y \approx 1.0 + 26.85 - 26.42 = 1.43 \mathrm{~m} \]
04

Compare with the Wall Height

Compare the calculated height \( y \approx 1.43 \mathrm{~m} \) with the wall height of \( 3.5 \mathrm{~m} \). Since \( 1.43 \mathrm{~m} < 3.5 \mathrm{~m} \), the ball does not clear the wall.

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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 this motion. It's a crucial component of projectile motion analysis. In any projectile problem, like the one involving the soccer ball, kinematics allows us to understand how the ball moves from point A to point B.
A key part of kinematics is determining the motion parameters such as displacement, speed, velocity, and acceleration.
  • Velocity in kinematics describes how fast and in which direction an object is moving.
  • Displacement is the overall change in position.
  • Acceleration refers to how the speed of the object changes over time.
By applying kinematic equations, we can predict the future position of the ball based on its initial velocity and other parameters like time and angle.
Vertical Motion
Vertical motion in projectile motion refers to the upward and downward movement of an object. It's affected by gravity, which slows down the object as it moves upward and speeds it up as it falls. The vertical component of motion can be analyzed using kinematic equations.
In this exercise, the vertical motion of the soccer ball is calculated using the initial vertical velocity and the time it takes to reach the wall. The equation used is: \[ y = y_0 + v_{0y} \cdot t_w - \frac{1}{2} g t_w^2 \]where:
  • \(y_0\) is the initial vertical position.
  • \(v_{0y}\) is the initial vertical velocity.
  • \(g\) is the acceleration due to gravity (9.8 m/s虏).
This equation helps determine the height of the ball when it strikes or passes by an obstacle, such as a wall.
Horizontal Motion
Horizontal motion in projectile problems deals with the movement of an object along the horizontal plane. In contrast to vertical motion, horizontal motion is unaffected by gravity and thus remains constant. This implies that the horizontal component of velocity remains steady as the object travels.
In the problem, we calculate the time it takes for the ball to reach the wall at a distance of 32 meters using the equation:
  • \[ t_w = \frac{x}{v_{0x}} \]
    • where:
    • \(x\) is the horizontal distance to the wall.
    • \(v_{0x}\) is the initial horizontal velocity.
    Once the time to reach the wall is known, it aids in further calculating the vertical position.
Trigonometric Components
The initial velocity of a projectile must be broken down into two components to fully understand its motion: horizontal and vertical. This breakdown is done using trigonometric functions based on the angle of launch.
In this exercise, the ball is kicked at an angle of 40掳 from the horizontal. The initial speed is 18 m/s. To find out how the ball will travel, we use:
  • Horizontal component \[ v_{0x} = v_0 \cdot \cos(40^\circ) \]
  • Vertical component \[ v_{0y} = v_0 \cdot \sin(40^\circ) \]
These components allow us to calculate separate motions and apply appropriate kinematic equations. Understanding these components is crucial in predicting the behavior of the projectile, like how high and far it will travel. By knowing these, we can solve for the time of flight, range, and maximum height of projectiles.

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Most popular questions from this chapter

A stone is thrown by aiming directly at the center \(P\) of a picture hanging on a wall. The stone leaves from the starting point horizontally with a speed of \(6.75 \mathrm{~m} / \mathrm{s}\) and strikes the target at point \(Q\), which is \(5.00 \mathrm{~cm}\) below \(P\). Find the horizontal distance between the starting point of the stone and the target.

Two seconds after being projected from ground level, a projectile is displaced \(40 \mathrm{~m}\) horizontally and \(58 \mathrm{~m}\) vertically above its launch point. What are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from the launch point?

A \(200 \mathrm{~m}\) wide river has a uniform flow speed of \(1.1 \mathrm{~m} / \mathrm{s}\) through a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of \(5.0 \mathrm{~m} / \mathrm{s}\) with respect to the water. There is a clearing on the north bank \(82 \mathrm{~m}\) upstream from a point directly opposite the clearing on the south bank. (a) In what direction must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank? (b) How long will the boat take to cross the river and land in the clearing?

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?

A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking \(2.50 \mathrm{~s}\). Then security agents appear, and the man runs as fast as he can back along the sidewalk to his starting point, taking \(10.0 \mathrm{~s}\). What is the ratio of the man's running speed to the sidewalk's speed?
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