Problem 64 - University Physics with Modern Physics

Question

A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0-m-high cliff. At the instant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 6.00 m/s. The woman runs in a straight line on level ground. Ignore air resistance on the ball. (a) At what angle above the horizontal should the ball be thrown so that the runner will catch it just before it hits the ground, and how far does she run before she catches the ball? (b) Carefully sketch the ball's trajectory as viewed by (i) a person at rest on the ground and (ii) the runner.

Step-by-Step Solution

Answer

(a) The angle is approximately 72.54°, and the runner covers a distance calculated by her speed and the time derived from the ball's fall. (b) Sketch depends on relative perspectives.

Step 1: Understand the Problem

We need to find at what angle the ball should be thrown such that it lands at the same position where the runner reaches after some time. The ball is thrown from the top of a cliff with an initial vertical speed and needs to meet the runner who starts running from the base of the cliff with constant speed.

Step 2: Ball Trajectory Analysis

Use the equations of motion to analyze the ball's trajectory. The initial speed is given as 20.0 m/s. Let the angle above the horizontal be \( \theta \). We have two components of the initial velocity: horizontal \( v_{0x} = v_0 \cos(\theta) \) and vertical \( v_{0y} = v_0 \sin(\theta) \).

Step 3: Determine Time of Flight

To determine when the ball hits the ground, use the vertical motion equation: \( y = h + v_{0y} t - \frac{1}{2}gt^2 \). Set \( y = 0 \) (ground level) and solve for \( t \):\[ 0 = 45 + 20\sin(\theta)t - \frac{1}{2}(9.8)t^2 \]Solving for \( t \) gives the time of flight equation in terms of \( \theta \).

Step 4: Calculate Distance Runner Covers

The runner's distance is calculated using \( d = vt \), where \( v = 6 \) m/s and \( t \) is derived from the ball hitting the ground. Set this distance equal to the horizontal distance the ball travels, \( d = v_{0x} t = 20\cos(\theta)t \), giving:\[ 6t = 20\cos(\theta)t \]From simplifying, \[ \frac{6}{20} = \cos(\theta) \].

Step 5: Solve for Angle \( \theta \)

Solve \( \cos(\theta) = 0.3 \) to find the angle \( \theta \):\[ \theta = \cos^{-1}(0.3) \approx 72.54^\circ \].

Step 6: Calculate the Run Distance

Use the runner's speed and time to determine how far she runs:\[ d = 6t \]With \( \theta \approx 72.54^\circ \), calculate \( t \) from the vertical motion equation and substitute it here to find \( d \).

Step 7: Sketch Trajectories

For part (b), sketch the parabola for a fixed observer that follows the general trajectory of projectiles. As the runner sees it (in her frame), the ball moves in a straight line path along a trajectory, due to matching horizontal motion.

Key Concepts

Trajectory Analysis

Trajectory analysis refers to the path that a projectile follows through space under the influence of gravitational forces. In this exercise, the ball is thrown from the top of a cliff which means its initial position is elevated, influencing its trajectory. Typically, a projectile follows a parabolic path. To analyze this path, we need to consider both the horizontal and vertical components of the initial velocity, as provided in the statement, and how they contribute to the overall motion of the projectile. The initial velocity of 20.0 m/s can be broken down into components based on the angle of projection. Understanding the projectile's trajectory is crucial because it helps us determine where and when the ball will land, allowing us to synchronize this with the runner's path.

Equations of Motion

Equations of motion describe how an object's velocity and position change over time due to external forces, such as gravity. For a projectile, we primarily use the following kinematic equations to determine its motion:

• Horizontal motion: - For any point in time, the horizontal distance can be described by the equation: \( x = v_{0x} t \).
• Vertical motion: - The vertical position is altered by gravity, given by the kinematic equation: \( y = h + v_{0y} t - \frac{1}{2}gt^2 \), where \( h \) is the initial height and \( g \) is the acceleration due to gravity.

These equations allow us to calculate how long the projectile will be in motion (time of flight) and how far it will travel horizontally when analyzed together.

Horizontal and Vertical Components

When breaking down projectile motion, it's essential to separate the initial velocity into horizontal and vertical components. The horizontal component \( v_{0x} = v_0 \cos(\theta) \) influences how far the projectile will travel, while the vertical component \( v_{0y} = v_0 \sin(\theta) \) affects the projectile's time in the air and the height it can reach. Both components are analyzed independently:

• Horizontal motion has no acceleration (ignoring air resistance), thus the velocity remains constant, impacting how far from the cliff the ball will move horizontally.
• Vertical motion is affected by gravity, which causes the ball to decelerate upward initially and accelerate downward thereafter, impacting when the ball will reach the ground.

Correctly calculating these components of motion is vital for solving for the angle \( \theta \) such that the runner intercepts the projectile at the precise location.

Time of Flight

The time of flight is the duration for which a projectile remains in motion from the launch until it lands back on the ground. To determine the time of flight, we solve the vertical motion equation by setting the final vertical position equal to zero (ground level): \( 0 = h + v_{0y} t - \frac{1}{2}gt^2 \). This quadratic equation can yield one or two solutions for \( t \), out of which the positive value of \( t \) is considered as it represents the realistic scenario (i.e., after the launch).In this example:

• Substituting in the values \( h = 45 \), \( v_{0y} = 20 \sin(\theta) \), and \( g = 9.8 \) m/s², you solve for \( t \).
• Incorporating \( \theta \) determined from the horizontal condition helps find the precise time for which the runner's path and the projectile coincide.

Understanding the time of flight helps us synchronize both the ball and the runner to meet just before impact.

Relative Motion Analysis

Relative motion analysis involves understanding how the motion of the projectile appears from different frames of reference. In this problem, we consider two viewpoints:

• From a stationary observer on the ground, the ball follows a typical parabolic path as expected from projectile motion.
• For the runner, who is moving at the same horizontal speed as the horizontal component of the projectile's motion, the ball appears to move vertically and in a straight path relative to her frame of motion.

To the runner, accounting for her speed of 6.00 m/s, the horizontal relative speed of the ball is zero. Consequently, she perceives the ball as moving directly towards her. Understanding relative motion is crucial in physics as it allows for explaining how different observers may conceptualize the same moving object in various ways.