Problem 45 - College Physics

Question

\(\bullet\) 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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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:\( v_{0x} = 12.0 \cos(40.0^{\circ}) \approx 9.19 \text{ m/s} \) \( v_{0y} = 12.0 \sin(40.0^{\circ}) \approx 7.71 \text{ m/s} \)

Step 3: 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 \( g = 9.8 \text{ m/s}^2 \):\( t = \frac{2 \times 7.71}{9.8} \approx 1.57 \text{ s} \)

Step 4: 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:\( R = 9.19 \times 1.57 \approx 14.42 \text{ m} \)

Step 5: 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:\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{30.0 - 14.42}{1.57} \approx 9.88 \text{ m/s} \]

Key Concepts

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.