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Chapter 2: Problem 76

At the beginning of a basketball game, a referee tosses the ball straight up with a speed of 4.6 \(\mathrm{m} / \mathrm{s}\) . A player cannot touch the ball until after it reaches its maximum height and begins to fall down. What is the minimum time that a player must wait before touching the ball?

Short Answer

Expert verified
The player must wait approximately 0.47 seconds.

Step by step solution

01

Identify the variables and given data

In this problem, we have the initial speed of the ball, \( v_i = 4.6 \, \text{m/s} \), and the acceleration due to gravity that will act on the ball, which is \( g = 9.8 \, \text{m/s}^2 \) downwards.
02

Set the final velocity at maximum height to zero

At the maximum height, the velocity of the ball becomes zero as it stops ascending and prepares to descend. Thus, the final velocity \( v_f = 0 \, \text{m/s} \).
03

Use the kinematic equation to solve for time

The kinematic equation connecting initial velocity, final velocity, acceleration, and time is \( v_f = v_i + at \). Rearranging for time \( t \), we find \( t = \frac{v_f - v_i}{a} \).
04

Substitute the known values into the equation

Substitute \( v_f = 0 \, \text{m/s}\), \( v_i = 4.6 \, \text{m/s} \), and \( a = -9.8 \, \text{m/s}^2 \) into the equation, giving us \( t = \frac{0 - 4.6}{-9.8} \).
05

Calculate the time

Calculate the time it takes for the ball to reach the maximum height: \( t = \frac{-4.6}{-9.8} \approx 0.47 \, \text{seconds} \).

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Key Concepts

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

Initial velocity
Initial velocity is the speed at which an object begins its motion. When a referee tosses a basketball upwards, the initial velocity is how fast it leaves the referee's hands. For this example, the initial velocity is given as 4.6 meters per second. This velocity impacts how high the ball will rise and for how long.

The greater the initial velocity, the higher and longer the ball will travel. It's the starting push that sets everything in motion. In the formula we used for this exercise, initial velocity appears as \( v_i \), which represents the ball鈥檚 speed at the start.
Acceleration due to gravity
Gravity is a force that pulls objects toward the Earth. It makes things fall when they're thrown upwards. Acceleration due to gravity is a constant value on Earth, approximately 9.8 meters per second squared. This value tells us how fast the speed of an object changes as it falls back towards the ground.

In the problem, gravity acts downward, opposite to the direction in which the ball is initially thrown. This is why we use a negative sign \(-9.8 \, \text{m/s}^2\) in our equations. This force will eventually stop the ball at its peak before pulling it back down.
Final velocity
Final velocity refers to the speed of an object at a specific point in time. In the context of this exercise, we're interested in the final velocity when the ball reaches its maximum height. At this point, the velocity is zero because the ball stops moving upwards before gravity pulls it back down.

This concept is crucial in calculations, as we can use it to determine the time taken to reach the maximum height. The formula \( v_f = 0 \, \text{m/s} \) allows us to isolate the time variable in kinematic equations.
Time of flight
Time of flight is the total time an object spends in the air, from launch to when it returns to the starting point. In this exercise, we only calculate the time to reach the maximum height, which is half the flight. We used the kinematic equation \( v_f = v_i + at \) to rearrange for time, giving us \( t = \frac{v_f - v_i}{a} \).

Substituting the values given and calculated (\( v_f = 0 \, \text{m/s}, \, v_i = 4.6 \, \text{m/s}, \, a = -9.8 \, \text{m/s}^2 \)), we found the time to be approximately 0.47 seconds. This duration tells us how long the player must wait before the ball starts descending, and it's an essential part of understanding projectile motion.

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Most popular questions from this chapter

A motorcycle has a constant acceleration of 2.5 \(\mathrm{m} / \mathrm{s}^{2} .\) Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from (a) 21 to \(31 \mathrm{m} / \mathrm{s},\) and \(\quad\) (b) 51 to 61 \(\mathrm{m} / \mathrm{s} ?\)

A jogger accelerates from rest to 3.0 \(\mathrm{m} / \mathrm{s}\) in 2.0 s. A car accelerates from 38.0 to 41.0 \(\mathrm{m} / \mathrm{s}\) also in 2.0 \(\mathrm{s}\) . (a) Find the acceleration (magnitude only) of the jogger. (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel farther than the jogger during the 2.0 \(\mathrm{s} ?\) If so, how much farther?

A football player, starting from rest at the line of scrimmage, accelerates along a straight line for a time of 1.5 s. Then, during a negligible amount of time, he changes the magnitude of his acceleration to a value of 1.1 \(\mathrm{m} / \mathrm{s}^{2}\) . With this acceleration, he continues in the same direction for another 1.2 \(\mathrm{s}\) , until he reaches a speed of 3.4 \(\mathrm{m} / \mathrm{s}\) . What is the value of his acceleration (assumed to be constant) during the initial 1.5 -s period?

A police car is traveling at a velocity of 18.0 \(\mathrm{m} / \mathrm{s}\) due north, when a car zooms by at a constant velocity of 42.0 \(\mathrm{m} / \mathrm{s}\) due north. After a reaction time of 0.800 \(\mathrm{s}\) s the policeman begins to pursue the speeder with an acceleration of 5.00 \(\mathrm{m} / \mathrm{s}^{2}\) . Including the reaction time, how long does it take for the police car to catch up with the speeder?

Before starting this problem, review Multiple-Concept Example \(6 .\) The left ventricle of the heart accelerates blood from rest to a velocity of \(+26 \mathrm{cm} / \mathrm{s}\) . (a) If the displacement of the blood during the acceleration is \(+2.0 \mathrm{cm},\) determine its acceleration (in \(\mathrm{cm} / \mathrm{s}^{2} )\) (b) How much time does blood take to reach its final velocity?
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