**Physics Problem - Baseball Throw Upward**A baseball is thrown directly upward with an initial velocity of \(75 \ \text{ft/sec}\) from an initial height of \(30 \ \text{ft}\). The velocity of the baseball, given by \(\nu(t) = -32t + 75\), where \(t\) is the number of seconds since the ball was released and \(\nu\) is in ft/sec.### Tasks:1. **Find the Function for Height**: Find the function, \(h\), that gives the height (in feet) of the baseball after \(t\) seconds if we know that at \( t = 0 \), the ball is 30 ft above the ground and the velocity of the baseball at any time \(t\) is given by \(\nu(t) = -32t + 75\).2. **Calculate Height After 2 Seconds**: What is the height of the baseball after \(2\) seconds of flight?3. **Time to Reach Highest Point**: After how many seconds does the ball reach its highest point? (Hint: The ball 'stops' for a moment before falling downward, \(\nu(t) = 0\)).4. **Maximum Height**: How high does the ball get at its highest point? (Use the time found in the previous step to compute this using the height function).5. **Time to Hit the Ground**: After how many seconds will the ball hit the ground? (The ball will hit the ground when the height \(h(t)\) becomes \(0\)).6. **Velocity on Ground Impact**: What is the ball’s velocity at the moment it hits the ground? (Use the time found in the previous step to compute the velocity when the ball hits the ground).For a deeper understanding, remember to derive from the motion equations considering gravity as the acceleration, which in this problem is \(-32 \ \text{ft/sec}^2\).### Mathematical Solutions & Explanations:a) To find the height function \(h(t)\): We start with the given velocity function, \(\nu(t) = -32t + 75\). Velocity, \(\nu(t)\), is the derivative of the height function, \(h(t)\): \[ \nu(t) = \frac{dh(t

Name: **Physics Problem - Baseball Throw Upward**A baseball is thrown direc
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Description: **Physics Problem - Baseball Throw Upward**A baseball is thrown directly upward with an initial velocity of \(75 \ \text{ft/sec}\) from an initial height of \(30 \ \text{ft}\). The velocity of the baseball, given by \(\nu(t) = -32t + 75\), where \(t

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7. Physics: A baseball is thrown directly upward with initial velocity of 75 ft/sec from an initial height of 30 ft. The velocity of the baseball is given by: v(t) = –32t + 75, wheret is the number of seconds since the ball was released and v is in ft/sec. a) Find the function h that gives the height (in feet) of the baseball after t seconds if we know that at t = 0, the ball is 30 ft above ground. b) what are the height and velocity of the baseball after 2 seconds of flight? c) After how many seconds does the ball reach its highest point? (hint: the ball "stops" for a moment before falling downward, v(t) = 0) %3D

7. Physics: A baseball is thrown directly upward with initial velocity of 75 ft/sec from an initial height of 30 ft. The velocity of the baseball is given by: v(t) = –32t + 75, wheret is the number of seconds since the ball was released and v is in ft/sec. a) Find the function h that gives the height (in feet) of the baseball after t seconds if we know that at t = 0, the ball is 30 ft above ground. b) what are the height and velocity of the baseball after 2 seconds of flight? c) After how many seconds does the ball reach its highest point? (hint: the ball "stops" for a moment before falling downward, v(t) = 0) %3D

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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Question

**Physics Problem - Baseball Throw Upward**

A baseball is thrown directly upward with an initial velocity of \(75 \ \text{ft/sec}\) from an initial height of \(30 \ \text{ft}\). The velocity of the baseball, given by \(\nu(t) = -32t + 75\), where \(t\) is the number of seconds since the ball was released and \(\nu\) is in ft/sec.

### Tasks:

1. **Find the Function for Height**:
Find the function, \(h\), that gives the height (in feet) of the baseball after \(t\) seconds if we know that at \( t = 0 \), the ball is 30 ft above the ground and the velocity of the baseball at any time \(t\) is given by \(\nu(t) = -32t + 75\).

2. **Calculate Height After 2 Seconds**:
What is the height of the baseball after \(2\) seconds of flight?

3. **Time to Reach Highest Point**:
After how many seconds does the ball reach its highest point? (Hint: The ball 'stops' for a moment before falling downward, \(\nu(t) = 0\)).

4. **Maximum Height**:
How high does the ball get at its highest point? (Use the time found in the previous step to compute this using the height function).

5. **Time to Hit the Ground**:
After how many seconds will the ball hit the ground? (The ball will hit the ground when the height \(h(t)\) becomes \(0\)).

6. **Velocity on Ground Impact**:
What is the ball’s velocity at the moment it hits the ground? (Use the time found in the previous step to compute the velocity when the ball hits the ground).

For a deeper understanding, remember to derive from the motion equations considering gravity as the acceleration, which in this problem is \(-32 \ \text{ft/sec}^2\).

### Mathematical Solutions & Explanations:

a) To find the height function \(h(t)\):

We start with the given velocity function, \(\nu(t) = -32t + 75\).

Velocity, \(\nu(t)\), is the derivative of the height function, \(h(t)\):

\[
\nu(t) = \frac{dh(t

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