7)If a ball is Ehrown into Ehe air with a velociby of sIftis, it's height in feet after t seconds isgiven by HCE1=516-16? The velocity after 2 second woula be:?
7)If a ball is Ehrown into Ehe air with a velociby of sIftis, it's height in feet after t seconds isgiven by HCE1=516-16? The velocity after 2 second woula be:?
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Educational Transcription:**
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**Problem 7: Motion of a Ball Thrown into the Air**
**Question:**
If a ball is thrown into the air with a velocity of 51 ft/s, its height in feet after \( t \) seconds is given by the equation:
\[ H(t) = 51t - 16t^2 \]
**Task:**
Calculate the velocity of the ball after 2 seconds.
**Graph Explanation:**
The graph on the page appears to be a hand-drawn sketch of a parabola, which represents the height \( H(t) \) of the ball over time \( t \). The x-axis likely represents the time in seconds, and the y-axis represents the height in feet. The shape of the parabola indicates that the height increases initially, reaches a peak, and then decreases as time progresses due to the effects of gravity.
To solve the problem:
1. **Find the derivative of \( H(t) \)**: The velocity of the ball as a function of time is the derivative of the height function \( H(t) \).
\[
H'(t) = \frac{d}{dt}(51t - 16t^2) = 51 - 32t
\]
2. **Calculate the velocity after 2 seconds**:
\[
H'(2) = 51 - 32(2) = 51 - 64 = -13 \text{ ft/s}
\]
Therefore, the velocity of the ball after 2 seconds is \(-13 \text{ ft/s}\), indicating that the ball is moving downward at this moment.
---
This transcription and explanation are aimed at aiding students in understanding the problem-solving process related to motion under the influence of gravity.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F08f6f3ae-62d0-4ecf-a7d1-a70a7054db3c%2F290874d5-01a6-4c24-bd4f-3fff74a1a19a%2F5w64tvc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Transcription:**
---
**Problem 7: Motion of a Ball Thrown into the Air**
**Question:**
If a ball is thrown into the air with a velocity of 51 ft/s, its height in feet after \( t \) seconds is given by the equation:
\[ H(t) = 51t - 16t^2 \]
**Task:**
Calculate the velocity of the ball after 2 seconds.
**Graph Explanation:**
The graph on the page appears to be a hand-drawn sketch of a parabola, which represents the height \( H(t) \) of the ball over time \( t \). The x-axis likely represents the time in seconds, and the y-axis represents the height in feet. The shape of the parabola indicates that the height increases initially, reaches a peak, and then decreases as time progresses due to the effects of gravity.
To solve the problem:
1. **Find the derivative of \( H(t) \)**: The velocity of the ball as a function of time is the derivative of the height function \( H(t) \).
\[
H'(t) = \frac{d}{dt}(51t - 16t^2) = 51 - 32t
\]
2. **Calculate the velocity after 2 seconds**:
\[
H'(2) = 51 - 32(2) = 51 - 64 = -13 \text{ ft/s}
\]
Therefore, the velocity of the ball after 2 seconds is \(-13 \text{ ft/s}\), indicating that the ball is moving downward at this moment.
---
This transcription and explanation are aimed at aiding students in understanding the problem-solving process related to motion under the influence of gravity.
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