Consider an object launched vertically upward from the top of a hill that is 144 ft above level with an initial velocity of 96 Alsec. Hs position would be given by hhl + 79b + 791-= (7 What is the average velocity over the first 3 seconds of flignt. Must calculate net distance traveled t=3 - t=0 and divide by t=3 s(3)- slo) - [-16(3)² + 96(3) +144] - [- 16(0)² + 96(0) + 144] . 3-0 144 3 48 ft/sec What is the instantaneos velocity at t=3 ? For example, compute average velocity from t- 3.99 to t=4.01 s(4.01)- s(3.99) = - 32 A/sec an estimate of velocity at t=4 4.01-3.99) average = s(t+At)-st). s(t+At) -s(t) ++At -t At lim (- 32t - 16 ot + 96) At >0 A reasonable velocity function of the projectile would be vlt)= 96-32t *derivative of -16t?+96t + 194 Extra Credit: How seconds until the object reaches maximum Bheight many and what is the maximum height?
Consider an object launched vertically upward from the top of a hill that is 144 ft above level with an initial velocity of 96 Alsec. Hs position would be given by hhl + 79b + 791-= (7 What is the average velocity over the first 3 seconds of flignt. Must calculate net distance traveled t=3 - t=0 and divide by t=3 s(3)- slo) - [-16(3)² + 96(3) +144] - [- 16(0)² + 96(0) + 144] . 3-0 144 3 48 ft/sec What is the instantaneos velocity at t=3 ? For example, compute average velocity from t- 3.99 to t=4.01 s(4.01)- s(3.99) = - 32 A/sec an estimate of velocity at t=4 4.01-3.99) average = s(t+At)-st). s(t+At) -s(t) ++At -t At lim (- 32t - 16 ot + 96) At >0 A reasonable velocity function of the projectile would be vlt)= 96-32t *derivative of -16t?+96t + 194 Extra Credit: How seconds until the object reaches maximum Bheight many and what is the maximum height?
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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![### Projectile Motion Analysis
**Problem Statement:**
Consider an object launched vertically upward from the top of a hill that is 144 feet above level with an initial velocity of 96 ft/sec. Its position is given by the function:
\[ s(t) = -16t^2 + 96t + 144 \]
**1. Average Velocity Calculation:**
What is the average velocity over the first 3 seconds of flight?
- **Formula:** \(\dfrac{s(t) - s(0)}{t - 0}\)
- For \( t = 3 \):
\[
\dfrac{[-16(3)^2 + 96(3) + 144] - [-16(0)^2 + 96(0) + 144]}{3} = \dfrac{144}{3} = 48 \text{ ft/sec}
\]
**2. Instantaneous Velocity at \( t = 3 \):**
To find an estimate of the instantaneous velocity at \( t = 3 \), compute the average velocity from \( t = 3.99 \) to \( t = 4.01 \):
\[
\text{Velocity} = \dfrac{s(4.01) - s(3.99)}{4.01 - 3.99} = 32 \text{ ft/sec}
\]
This provides an estimate for the velocity at \( t = 4 \).
**Formula for Instantaneous Velocity:**
\[
\lim_{\Delta t \to 0} \dfrac{s(t + \Delta t) - s(t)}{\Delta t}
\]
A reasonable velocity function derived from the position function is:
\[ v(t) = 96 - 32t \]
This is obtained from the derivative of \( s(t) = -16t^2 + 96t + 144 \).
**Extra Credit:**
How many seconds until the object reaches maximum height, and what is the maximum height?
---
This exploration covers fundamental concepts in projectile motion, average and instantaneous velocity calculations, and the use of derivatives in physics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff26050af-8cf2-44aa-a499-323c21b5df7b%2F133b5d92-85ca-4183-87b2-e77b2f9c2313%2F60fb09g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Projectile Motion Analysis
**Problem Statement:**
Consider an object launched vertically upward from the top of a hill that is 144 feet above level with an initial velocity of 96 ft/sec. Its position is given by the function:
\[ s(t) = -16t^2 + 96t + 144 \]
**1. Average Velocity Calculation:**
What is the average velocity over the first 3 seconds of flight?
- **Formula:** \(\dfrac{s(t) - s(0)}{t - 0}\)
- For \( t = 3 \):
\[
\dfrac{[-16(3)^2 + 96(3) + 144] - [-16(0)^2 + 96(0) + 144]}{3} = \dfrac{144}{3} = 48 \text{ ft/sec}
\]
**2. Instantaneous Velocity at \( t = 3 \):**
To find an estimate of the instantaneous velocity at \( t = 3 \), compute the average velocity from \( t = 3.99 \) to \( t = 4.01 \):
\[
\text{Velocity} = \dfrac{s(4.01) - s(3.99)}{4.01 - 3.99} = 32 \text{ ft/sec}
\]
This provides an estimate for the velocity at \( t = 4 \).
**Formula for Instantaneous Velocity:**
\[
\lim_{\Delta t \to 0} \dfrac{s(t + \Delta t) - s(t)}{\Delta t}
\]
A reasonable velocity function derived from the position function is:
\[ v(t) = 96 - 32t \]
This is obtained from the derivative of \( s(t) = -16t^2 + 96t + 144 \).
**Extra Credit:**
How many seconds until the object reaches maximum height, and what is the maximum height?
---
This exploration covers fundamental concepts in projectile motion, average and instantaneous velocity calculations, and the use of derivatives in physics.
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