/*! 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 85 Wrongly called for a foul, an an... [FREE SOLUTION] | 91Ó°ÊÓ

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Wrongly called for a foul, an angry basketball player throws the ball straight down to the floor. If the ball bounces straight up and returns to the floor \(2.8 \mathrm{s}\) after first striking it, what was the ball's greatest height above the floor?

Short Answer

Expert verified
The ball's greatest height above the floor was approximately 9.61 meters.

Step by step solution

01

Recognize the Problem Type

This is a motion problem dealing with free fall under gravity. We have to calculate the maximum height the ball reaches after bouncing back from the floor.
02

Determine the Time to Reach Maximum Height

The total time for the ball's journey is given as 2.8 seconds. We know that this time includes the time taken to reach the maximum height and the time taken to return to the floor. As the motion is symmetric, it takes half of the total time to reach the highest point. Therefore, the time to reach the maximum height is \( t = \frac{2.8}{2} = 1.4 \, \mathrm{s} \).
03

Use the Kinematic Equation to Find Initial Velocity

We can use the kinematic equation: \[ v_f = v_i + a t \] where: - \( v_f = 0 \) (velocity at the highest point), - \( a = -9.8 \, \mathrm{m/s^2} \) (acceleration due to gravity). Rearranging to find \( v_i \) (initial velocity) gives \[ 0 = v_i - 9.8 \times 1.4 \] \[ v_i = 9.8 \times 1.4 = 13.72 \, \mathrm{m/s} \].
04

Calculate the Maximum Height Using Initial Velocity

We now use another kinematic equation to find the maximum height: \[ v_f^2 = v_i^2 + 2 a h \] where \( v_f = 0 \) at the maximum height. Solving for height \( h \): \[ 0 = (13.72)^2 + 2(-9.8)h \] \[ h = \frac{(13.72)^2}{2 \times 9.8} \approx 9.61 \, \mathrm{meters} \].

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Key Concepts

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

Free Fall
Free fall is a straightforward concept in physics that refers to any motion of a body where gravity is the only force acting upon it. When you drop something from a height, or when an object is tossed upwards, it follows a free fall. In this scenario, the basketball, after being thrown downward, bounces back up, undergoing free fall due to gravity alone. While moving upward or downward, air resistance can typically be ignored in an ideal physics problem, making calculations based solely on gravitational acceleration.
Always remember, in a free fall, gravity accelerates the object downward at a constant rate, approximately \(-9.8 \ m/s^2\). This simplicity allows us to apply kinematic equations effectively to solve these problems.
Maximum Height
Understanding the maximum height is crucial in motion problems, especially those involving projectile motion or free fall. The maximum height is the highest point an object reaches during its motion before it starts descending again. When solving for maximum height, you consider the point where the upward velocity becomes zero.
In this case, the basketball bounces back and reaches its greatest height before descending due to gravitational pull. We can use the equation \(v_f^2 = v_i^2 + 2ah\) to find this height. Here, \(v_f = 0\) because that's the velocity at the highest point and \(a = -9.8 \ m/s^2\). Solving for the height gives us a clearer view of how high the ball traveled post-bounce.
Initial Velocity
Initial velocity is the speed at which an object begins its journey, either from rest or another state of motion. In our case, we're determining how fast the basketball leaves the ground post-bounce. This velocity is crucial because it sets the ball on its path upward until gravity slows it to a halt.
The equation used to find initial velocity here is \(v_f = v_i + at\), where \(v_f\) is zero at the maximum height because the ball stops momentarily there. Recalculating gives \(v_i = 9.8 \times 1.4 = 13.72 \ m/s\). Understanding initial velocity helps in predicting how high an object will reach when thrown upwards.
Kinematic Equations
In physics, kinematic equations are the powerful, universal tools used to analyze motion. They relate the five key variables in motion: displacement, initial velocity, final velocity, acceleration, and time.
For our exercise, the kinematic equations helped solve the problem by relating these variables in a form we could easily manipulate: - For velocity: \(v_f = v_i + at\)- For maximum height: \(v_f^2 = v_i^2 + 2ah\)
These equations allow any unknown variable, such as height or initial velocity, to be determined, provided enough other variables are known. They simplify complex motion scenarios by creating a structured approach to problem-solving, highlighting the relationships between motion parameters.

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Most popular questions from this chapter

Bill steps off a 3.0 -m-high diving board and drops to the water below. At the same time, Ted jumps upward with a speed of \(4.2 \mathrm{m} / \mathrm{s}\) from a \(1.0-\mathrm{m}\) -high diving board. Choosing the origin to be at the water's surface, and upward to be the positive \(x\) direction, write \(x\) -versus-t equations of motion for both Bill and Ted.

As a train accelerates away from a station, it reaches a specd of \(4.7 \mathrm{m} / \mathrm{s}\) in \(5.0 \mathrm{s}\). If the train's acceleration remains constant. what is its speed after an additional 6.0 s has elapsed?

A model rocket rises with constant acceleration to a height of \(3.2 \mathrm{m},\) at which point its speed is \(26.0 \mathrm{m} / \mathrm{s} .\) (a) How much time does it take for the rocket to reach this height? (b) What was the magnitude of the rocket's acceleration? (c) Find the height and speed of the rocket \(0.10 \mathrm{s}\) after launch.

You drop a ski glove from a height \(h\) onto fresh snow, and it sinks to a depth \(d\) before coming to rest. (a) In terms of \(g\) and \(h,\) what is the speed of the glove when it reaches the snow? (b) What are the magnitude and direction of the glove's acceleration as it moves through the snow, assuming it to be constant? Give your answer in terms of \(g, h,\) and \(d\).

A boat is cruising in a straight line at a constant speed of \(2.6 \mathrm{m} / \mathrm{s}\) when it is shifted into neutral. After coasting \(12 \mathrm{m}\) the engine is engaged again, and the boat resumes cruising at the reduced constant speed of \(1.6 \mathrm{m} / \mathrm{s}\). Assuming constant acceleration while coasting. (a) how long did it take for the boat to coast the \(12 \mathrm{m} ?\) (b) What was the boat's acceleration while it was coasting? \((c)\) When the boat had coasted for \(6.0 \mathrm{m},\) was its speed \(2.1 \mathrm{m} / \mathrm{s},\) more than \(2.1 \mathrm{m} / \mathrm{s},\) or less than \(2.1 \mathrm{m} / \mathrm{s}\) ? Explain.

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