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Chapter 2: Problem 47

A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height. Find (a) its initial velocity and (b) the height it reaches.

Short Answer

Expert verified
The initial velocity of the ball is approximately \(29.4 m/s\) and it reaches a maximum height of approximately \(44.1 m\).

Step by step solution

01

Find the initial velocity using the equation of motion

To calculate the initial velocity, use the equation of motion \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is time. Since at maximum height, the final velocity is 0, and the acceleration due to gravity is \( -9.8 m/s^2\), the time is given as 3.00 seconds. The equation becomes: \(0 = u - 9.8 * 3.00\). Solving for \(u\) will give us the initial velocity.
02

Calculate the initial velocity

Use algebra to solve the equation from Step 1 for \(u\). That is, \(u = 9.8 * 3.00\). By calculating this, you will obtain the initial velocity as \(29.4 m/s\).
03

Calculate the maximum height using the motion equation

Now that you have the initial velocity, solve for the maximum height using the motion equation \(h = ut + \frac{1}{2}at^2\), where \(h\) is the height, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is time. Plug the values \(u = 29.4 m/s\), \(a = -9.8 m/s^2\), and \(t = 3.00 s\) into the equation.
04

Solve for the maximum height

By substituting the given values into the equation from Step 3, \(h = 29.4 * 3.00 +\frac{1}{2}*(-9.8)*(3.00)^2 \), you can calculate the maximum height that the ball reaches.

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Key Concepts

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

Initial Velocity Calculation
When a baseball is hit and travels straight upward, the concept of initial velocity is pivotal. In projectile motion, the initial velocity is the speed at which the object starts its journey upward. For this exercise, we use the equation of motion:
  • \(v = u + at\)
Here's what these symbols represent:
  • \(v\): final velocity
  • \(u\): initial velocity
  • \(a\): acceleration
  • \(t\): time
When the ball reaches its peak, its final velocity, \(v\), is 0, meaning it momentarily stops before descending. With gravitational acceleration \(a\) being \(-9.8 \ m/s^2\) (since gravity pulls the ball down), and \(t = 3.00\) seconds, the equation simplifies to
  • \(0 = u - 9.8 \cdot 3.00\)
By solving, we find \(u = 29.4\) m/s. This initial velocity sets the ball in motion for its upwards journey.
Equations of Motion
Equations of motion are vital for solving any problem related to how objects move. These equations link different quantities such as velocity, acceleration, and displacement.They are used throughout physics to describe linear motion, including projectile motion. In our case, we see the equation \(v = u + at\) explaining the change between initial and final velocity under constant acceleration.Another important equation is used to find height:
  • \(h = ut + \frac{1}{2}at^2\)

Understanding the Variables

The variables stand for:
  • \(h\): height reached
  • \(u\): initial velocity
  • \(a\): acceleration
  • \(t\): time
During the ball's ascent in our problem, gravity (\(-9.8 \ m/s^2\)) acts as a negative acceleration because it opposes the ball鈥檚 movement. This provides the necessary information to calculate how high the baseball travels.
Maximum Height Calculation
Once we've determined the initial velocity, the maximum height is our next focus. This is the highest point the baseball reaches during its flight. Using the equation:
  • \(h = ut + \frac{1}{2}at^2\)
we can substitute the known values:
  • \(u = 29.4 \ m/s\)
  • \(a = -9.8 \ m/s^2\)
  • \(t = 3.00 \ s\)
The calculation becomes straightforward:
  • \( h = 29.4 \times 3.00 + \frac{1}{2} \times (-9.8) \times (3.00)^2 \)
Upon solving, we find the maximum height the ball achieves. Understanding this concept also involves recognizing how any object kicked or thrown upwards will gracefully slow down, stop, and then begin accelerating downwards due to gravity.

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Most popular questions from this chapter

The yellow caution light on a traffic signal should stay on long enough to allow a driver to either pass through the intersection or safely stop before reaching the intersection. A car can stop if its distance from the intersection is greater than the stopping distance found in the previous problem. If the car is less than this stopping distance from the intersection, the yellow light should stay on long enough to allow the car to pass entirely through the intersection. (a) Show that the yellow light should stay on for a time interval.$$\Delta t_{\text {light }}=\Delta t_{r}-\left(v_{0} / 2 a\right)+\left(s_{i} / v_{0}\right)$$,where \(\Delta t_{r}\) is the driver's reaction time, \(v_{0}\) is the velocity of the car approaching the light at the speed limit, \(a\) is the braking acceleration, and \(s_{i}\) is the width of the intersection. (b) As city traffic planner, you expect cars to approach an intersection \(16.0 \mathrm{m}\) wide with a speed of \(60.0 \mathrm{km} / \mathrm{h} .\) Be cautious and assume a reaction time of 1.10 s to allow for a driver's indecision. Find the length of time the yellow light should remain on. Use a braking acceleration of \(-2.00 \mathrm{m} / \mathrm{s}^{2}\).

A jet plane lands with a speed of \(100 \mathrm{m} / \mathrm{s}\) and can accelerate at a maximum rate of \(-5.00 \mathrm{m} / \mathrm{s}^{2}\) as it comes to rest.(a) From the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest? (b) Can this plane land on a small tropical island airport where the runway is \(0.800 \mathrm{km}\) long?

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by \(v=\left(-5.00 \times 10^{7}\right) t^{2}+\) \(\left(3.00 \times 10^{5}\right) t,\) where \(v\) is in meters per second and \(t\) is in seconds. The acceleration of the bullet just as it leaves the barrel is zero. (a) Determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel. (b) Determine the length of time the bullet is accelerated. (c) Find the speed at which the bullet leaves the barrel. (d) What is the length of the barrel?

A woman is reported to have fallen 144 ft from the 17 th floor of a building, landing on a metal ventilator box, which she crushed to a depth of 18.0 in. She suffered only minor injuries. Neglecting air resistance, calculate (a) the speed of the woman just before she collided with the ventilator, (b) her average acceleration while in contact with the box, and (c) the time it took to crush the box.

A golf ball is released from rest from the top of a very tall building. Neglecting air resistance, calculate (a) the position and (b) the velocity of the ball after \(1.00,2.00,\) and \(3.00 \mathrm{s}\).
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