A projectile is launched straight up in the air. Its height (in feet) t seconds after launch is given by the function f(t)= - 16t² +423t+6. Find its height when it is falling at 269 feet/second. C... Its height is If necessary, round to two decimal places. Do not include units.
A projectile is launched straight up in the air. Its height (in feet) t seconds after launch is given by the function f(t)= - 16t² +423t+6. Find its height when it is falling at 269 feet/second. C... Its height is If necessary, round to two decimal places. Do not include units.
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 Problem
A projectile is launched straight up in the air. Its height (in feet) \( t \) seconds after the launch is given by the function:
\[ f(t) = -16t^2 + 423t + 6 \]
Find its height when it is falling at 269 feet/second.
---
#### Solution:
We'll first need to determine at what time \( t \) the velocity of the projectile is \( -269 \) feet/second (negative because it is falling).
To do that, we differentiate the height function with respect to \( t \) to find the velocity function:
\[ f'(t) = \frac{d}{dt} (-16t^2 + 423t + 6) \]
\[ f'(t) = -32t + 423 \]
Now, set the velocity equal to \(-269\) feet/second and solve for \( t \):
\[ -32t + 423 = -269 \]
Rearrange to solve for \( t \):
\[ -32t = -269 - 423 \]
\[ -32t = -692 \]
\[ t = \frac{-692}{-32} \]
\[ t = 21.625 \text{ seconds} \]
Now, substitute \( t = 21.625 \) back into the original height function to find the height at this time:
\[ f(21.625) = -16(21.625)^2 + 423(21.625) + 6 \]
Calculate the value:
\[ f(21.625) = -16(467.640625) + 9142.875 + 6 \]
\[ f(21.625) = -7482.25 + 9142.875 + 6 \]
\[ f(21.625) = 1666.625 + 6 \]
\[ f(21.625) = 1666.63 \]
Therefore, the height of the projectile when it is falling at 269 feet/second is approximately \( 1666.63 \) feet (rounded to two decimal places).
**Final Answer:**
Its height is \( 1666.63 \)
> Please note: If necessary, round to two decimal places. Do not include units in the answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3186549c-b7ff-4372-aaad-030b5790f577%2Fac7a10bb-4215-4cdc-9c52-092281ecb7b1%2F92ap5x_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Projectile Motion Problem
A projectile is launched straight up in the air. Its height (in feet) \( t \) seconds after the launch is given by the function:
\[ f(t) = -16t^2 + 423t + 6 \]
Find its height when it is falling at 269 feet/second.
---
#### Solution:
We'll first need to determine at what time \( t \) the velocity of the projectile is \( -269 \) feet/second (negative because it is falling).
To do that, we differentiate the height function with respect to \( t \) to find the velocity function:
\[ f'(t) = \frac{d}{dt} (-16t^2 + 423t + 6) \]
\[ f'(t) = -32t + 423 \]
Now, set the velocity equal to \(-269\) feet/second and solve for \( t \):
\[ -32t + 423 = -269 \]
Rearrange to solve for \( t \):
\[ -32t = -269 - 423 \]
\[ -32t = -692 \]
\[ t = \frac{-692}{-32} \]
\[ t = 21.625 \text{ seconds} \]
Now, substitute \( t = 21.625 \) back into the original height function to find the height at this time:
\[ f(21.625) = -16(21.625)^2 + 423(21.625) + 6 \]
Calculate the value:
\[ f(21.625) = -16(467.640625) + 9142.875 + 6 \]
\[ f(21.625) = -7482.25 + 9142.875 + 6 \]
\[ f(21.625) = 1666.625 + 6 \]
\[ f(21.625) = 1666.63 \]
Therefore, the height of the projectile when it is falling at 269 feet/second is approximately \( 1666.63 \) feet (rounded to two decimal places).
**Final Answer:**
Its height is \( 1666.63 \)
> Please note: If necessary, round to two decimal places. Do not include units in the answer.
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