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(II) A ball player catches a ball \(3.2 \mathrm{~s}\) after throwing it vertically upward. With what speed did he throw it, and what height did it reach?

Short Answer

Expert verified
The initial throw speed was 15.68 m/s, and the maximum height was 12.55 m.

Step by step solution

01

Identify Given Values

We are given that the total time of flight of the ball is \(3.2\) seconds. This means the ball travels up and comes down in this time frame. Since the motion is symmetric, the time to reach the maximum height is half of the total time, which is \(1.6\) seconds.
02

Break Down the Motion

When the ball is thrown vertically upward and reaches its highest point, its velocity becomes \(0\) m/s at the maximum height. We can use this phase to calculate the initial velocity using the kinematic equation for uniformly accelerated motion.
03

Use Kinematic Equation for Initial Velocity

Use the equation \(v = u + at\) to find the initial velocity \(u\). Here, \(v=0\) m/s (velocity at the peak), \(a = -9.8\, \text{m/s}^2\) (acceleration due to gravity, negative as it is acting downwards), and \(t = 1.6\) s.Substituting the values, we have:\[0 = u - 9.8\times1.6\]Solving this gives: \(u = 15.68\,\text{m/s}\).
04

Calculate Maximum Height

Use the equation \(v^2 = u^2 + 2as\) to find the maximum height \(s\). Here, \(v = 0\) m/s and \(a = -9.8\,\text{m/s}^2\).Substituting the initial velocity \(u = 15.68\,\text{m/s}\) we obtained before, we get:\[0 = (15.68)^2 + 2(-9.8)s\]Solving for \(s\) gives:\[0 = 245.8624 - 19.6s\]\[19.6s = 245.8624\]\[s = 12.55 \,\text{meters}\].

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Key Concepts

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

Kinematic Equations
In physics, kinematic equations are a set of four equations that can describe the motion of an object under constant acceleration. This is particularly useful in understanding projectile motion. These equations allow us to relate the variables of motion such as initial velocity, final velocity, acceleration, time, and displacement.

You can use these equations to solve many real-world problems involving motion, like the exercise where a ball is thrown vertically. The two primary equations used here include:
  • \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time taken.
  • \(v^2 = u^2 + 2as\), where \(s\) is the displacement.
For the ball player exercise, the first equation helps calculate the initial velocity at which the ball is thrown. The second equation determines the maximum height the ball reaches. Understanding how to manipulate these equations is key, providing a thorough understanding of projectile motion.
Initial Velocity Calculation
Determining the initial velocity of a projectile is crucial in predicting its motion. In the given exercise, we find the initial velocity of the ball using one of the kinematic equations.
Since the ball is thrown vertically upward, it moves against gravity. At the maximum height, its velocity becomes zero. This understanding helps us set up the equation:
  • \(v = u + at\)
At the peak, \(v = 0\) and \(a = -9.8 \, \text{m/s}^2\), which is due to the downward acceleration of gravity. The time to reach the maximum height is half of the total flight time, so \(t = 1.6 \, \text{seconds}\). Substituting these values in, we solve for \(u\) to find that the ball's initial speed was \(15.68 \, \text{m/s}\).
This calculation tells us the speed needed to throw the ball such that it reaches its peak height precisely as required by the exercise.
Free Fall
Free fall describes the motion of an object under the influence of gravitational force only. In the context of the exercise, the ball's motion after being thrown vertically upwards can be divided into two parts: ascending and descending. Both of these parts are influenced by gravity without any other force acting on the ball.
During the ascent, the ball slows down until it reaches its maximum height, regulated by the Earth's gravitational pull with an acceleration of \(-9.8\, \text{m/s}^2\).
Once this maximum height is reached and its velocity hits zero, the ball begins to fall back down. This part of the motion is considered true free fall because the ball's descent is now only affected by gravity as it accelerates back down towards its starting point. Understanding free fall is essential for comprehending how objects move under the singular force of gravity, giving insight into the ball's trajectory in the given exercise.

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Most popular questions from this chapter

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between stops. The more stops there are, the slower the train's average speed. To get an idea of this problem, calculate the time it takes a train to make a \(9.0-\mathrm{km}\) trip in two situations: \((a)\) the stations at which the trains must stop are \(1.8 \mathrm{~km}\) apart (a total of 6 stations, including those at the ends); and (b) the stations are \(3.0 \mathrm{~km}\) apart (4 stations total). Assume that at each station the train accelerates at a rate of \(1.1 \mathrm{~m} / \mathrm{s}^{2}\) until it reaches \(95 \mathrm{~km} / \mathrm{h},\) then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at \(-2.0 \mathrm{~m} / \mathrm{s}^{2} .\) Assume it stops at each intermediate station for \(22 \mathrm{~s}\).

A fugitive tries to hop on a freight train traveling at a constant speed of \(5.0 \mathrm{~m} / \mathrm{s}\). Just as an empty box car passes him, the fugitive starts from rest and accelerates at \(a=1.2 \mathrm{~m} / \mathrm{s}^{2}\) to his maximum speed of \(6.0 \mathrm{~m} / \mathrm{s} .(a)\) How long does it take him to catch up to the empty box car? (b) What is the distance traveled to reach the box car?

(II) A car slows down uniformly from a speed of \(18.0 \mathrm{~m} / \mathrm{s}\) to rest in 5.00 s. How far did it travel in that time?

(II) Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of \(1.5 \mathrm{~m}\) above the ground (Fig. \(2-45) .\) When you quickly turn off the nozzle, you hear the water striking the ground next to you for another \(2.0 \mathrm{~s}\). What is the water speed as it leaves the nozzle?

(II) On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about \(0.28 \mu \mathrm{m} .\) A CD player's readout laser scans along the spiral's sequence of bits at a constant speed of about \(1.2 \mathrm{~m} / \mathrm{s}\) as the CD spins. (a) Determine the number \(N\) of digital bits that a CD player reads every second. (b) The audio information is sent to each of the two loudspeakers 44,100 times per second. Each of these samplings requires 16 bits and so one would (at first glance) think the required bit rate for a CD player is \(N_{0}=2\left(44,100 \frac{\text { samplings }}{\text { second }}\right)\left(16 \frac{\text { bits }}{\text { sampling }}\right)=1.4 \times 10^{6} \frac{\text { bits }}{\text { second }}\) where the 2 is for the 2 loudspeakers (the 2 stereo channels). Note that \(N_{0}\) is less than the number \(N\) of bits actually read per second by a CD player. The excess number of bits \(\left(=N-N_{0}\right)\) is needed for encoding and error-correction. What percentage of the bits on a \(\mathrm{CD}\) are dedicated to encoding and error-correction?

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