/*! 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 54 A baseball is hit so that it tra... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The ball's initial velocity is \(29.4 \,\mathrm{m/s}\), and it reaches a height of \(44.1 \,\mathrm{m}\).

Step by step solution

01

Identify the knowns and unknowns

The knowns in this problem are the acceleration due to gravity \(a = -9.8 \,\mathrm{m/s^{2}}\), the final velocity at the maximum height \(v_f = 0 \,\mathrm{m/s}\), and the time taken to reach the maximum height \(t = 3.0 \,\mathrm{s}\). The unknowns are the initial velocity (\(v_i\)) and the maximum height reached (\(h\)).
02

Apply the Equation of motion to find Initial Velocity

Here we can use the kinematic equation \(v_f = v_i+ at\) where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time to solve for the initial velocity. Substituting in the given values we get \(0 = v_i - (9.8 \,\mathrm{m/s^{2}})(3.0 \,\mathrm{s})\). Solving for \(v_i\) gives \(v_i = 29.4 \,\mathrm{m/s}\).
03

Apply another equation of motion to find Height

Now we can use the height kinematic equation \(h = v_i t + 0.5at^{2}\). Substituting in the given values we get \(h = (29.4 \,\mathrm{m/s})(3.0 \,\mathrm{s}) + 0.5(-9.8 \,\mathrm{m/s^{2}})(3.0 \,\mathrm{s})^{2}\). Solving for \(h\) gives us \(h = 44.1 \,\mathrm{m}\).

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Key Concepts

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

Initial Velocity
When discussing the motion of objects like a baseball being hit straight up, the initial velocity is a crucial factor. It determines how quickly an object is launched from the start.
In kinematics, the initial velocity often plays a significant role in predicting future motion. For the baseball problem, we are interested in finding how quickly the ball was moving right after being hit by the bat.
To find the initial velocity, we utilize the kinematic equation:
  • \(v_f = v_i + at\)
- Here, \(v_f\) is the final velocity (0 m/s at the maximum height), \(v_i\) is what we're solving for, \(a\) is the acceleration due to gravity (\(-9.8 \,\mathrm{m/s^2}\)), and \(t\) is the time (3 seconds).
- Plugging in these values, we rearrange the equation to solve for \(v_i\): \[v_i = v_f - at = 0 - (-9.8 \,\mathrm{m/s^2}) \times 3.0 \,\mathrm{s} = 29.4 \,\mathrm{m/s}\\] Hence, the initial velocity of the baseball is \(29.4 \,\mathrm{m/s}\).
Maximum Height
Achieving the maximum height is the point when the baseball stops moving upward and is about to start its descent. At this peak, its velocity becomes zero. Calculating this height gives insight into how "high" the ball actually travels.
To find the maximum height, we apply another important equation from kinematics:
  • \(h = v_i t + 0.5 a t^2\)
- We use the previously calculated initial velocity \(29.4 \,\mathrm{m/s}\), time \(3.0 \,\mathrm{s}\), and known acceleration \(-9.8 \,\mathrm{m/s^2}\).
- Substituting these values, the equation becomes:\[h = (29.4 \,\mathrm{m/s}) \times 3.0 \,\mathrm{s} + 0.5 \times (-9.8 \,\mathrm{m/s^2}) \times (3.0 \,\mathrm{s})^2\\]After performing the arithmetic, we find that the maximum height \(h\) is \(44.1 \,\mathrm{m}\). Therefore, the baseball reaches a height of \(44.1 \,\mathrm{m}\) before reversing its motion.
Acceleration Due to Gravity
The acceleration due to gravity is a constant that affects all objects moving under the Earth's gravitational field. It is a downward force that pulls objects towards the center of the Earth, measured as \(9.8 \,\mathrm{m/s^2}\).
In the context of our problem, this value is crucial as it dictates the motion of the baseball after being hit.
  • The negative sign in \(-9.8 \,\mathrm{m/s^2}\) indicates that gravity acts in the opposite direction to the initial upward motion.
Understanding gravity's role helps explain why objects slow down as they reach their highest point and eventually start to fall. Thus, it is this constant acceleration that influences the ball's upward motion, causing it to decelerate until it reaches the maximum height where its velocity is zero. This understanding allows us to predict the ball's trajectory and height effectively.

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Most popular questions from this chapter

A model rocket is launched straight upward with an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\). It accelerates with a constant upward acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) until its engines stop at an altitude of \(150 \mathrm{~m}\). (a) What can you say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air?

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