A stone is thrown with an initial velocity of 35 ft/s from the edge of a bridge that is 47 ft above the ground. The height of this stone above the ground t seconds after it is thrown is f(t) = - 16t + 35t + 47. If a second stone is thrown from the ground, then its height above the ground after t seconds is given by g(t) = – 16t + vot, where vo is the initial velocity of the second stone. Determine the value of vo such that the two stones reach the same high point. When an object that is thrown upwards reaches its highest point (just before it starts to fall back to the ground), its will be zero. Therefore, to find the maximum height of the object thrown from the bridge, use the equation, The initial velocity of the second stone would need to be vo (Do not round until the final answer. Then round to one decimal place.)
A stone is thrown with an initial velocity of 35 ft/s from the edge of a bridge that is 47 ft above the ground. The height of this stone above the ground t seconds after it is thrown is f(t) = - 16t + 35t + 47. If a second stone is thrown from the ground, then its height above the ground after t seconds is given by g(t) = – 16t + vot, where vo is the initial velocity of the second stone. Determine the value of vo such that the two stones reach the same high point. When an object that is thrown upwards reaches its highest point (just before it starts to fall back to the ground), its will be zero. Therefore, to find the maximum height of the object thrown from the bridge, use the equation, The initial velocity of the second stone would need to be vo (Do not round until the final answer. Then round to one decimal place.)
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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![**Problem Statement:**
A stone is thrown with an initial velocity of 35 ft/s from the edge of a bridge that is 47 ft above the ground. The height of this stone above the ground \( t \) seconds after it is thrown is \( f(t) = -16t^2 + 35t + 47 \).
If a second stone is thrown from the ground, then its height above the ground after \( t \) seconds is given by \( g(t) = -16t^2 + v_0t \), where \( v_0 \) is the initial velocity of the second stone. Determine the value of \( v_0 \) such that the two stones reach the same high point.
**Concept Explanation:**
When an object that is thrown upwards reaches its highest point (just before it starts to fall back to the ground), its velocity will be zero.
To find the maximum height of the object thrown from the bridge, use the following kinematic equation:
\[ v = u + at \]
**Goal:**
Determine the initial velocity \( v_0 \) of the second stone.
**Solution:**
The initial velocity of the second stone \( v_0 \) would need to be:
\[ v_0 = \_\_\_\_ \]
(Do not round until the final answer. Then round to one decimal place.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3b13cfd-0bfc-4fad-ae09-6a6a47693799%2Fb9f3b91b-ee9f-4044-b250-1de2697687e1%2Fuc9bks_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A stone is thrown with an initial velocity of 35 ft/s from the edge of a bridge that is 47 ft above the ground. The height of this stone above the ground \( t \) seconds after it is thrown is \( f(t) = -16t^2 + 35t + 47 \).
If a second stone is thrown from the ground, then its height above the ground after \( t \) seconds is given by \( g(t) = -16t^2 + v_0t \), where \( v_0 \) is the initial velocity of the second stone. Determine the value of \( v_0 \) such that the two stones reach the same high point.
**Concept Explanation:**
When an object that is thrown upwards reaches its highest point (just before it starts to fall back to the ground), its velocity will be zero.
To find the maximum height of the object thrown from the bridge, use the following kinematic equation:
\[ v = u + at \]
**Goal:**
Determine the initial velocity \( v_0 \) of the second stone.
**Solution:**
The initial velocity of the second stone \( v_0 \) would need to be:
\[ v_0 = \_\_\_\_ \]
(Do not round until the final answer. Then round to one decimal place.)
![**Understanding Profit Functions in Economics**
In this section, we explore how to calculate the profit function \( P(x) \), given cost and price functions. Let's delve into the provided formulas and examples.
**Problem Statement:**
Let \( C(x) \) represent the cost of producing \( x \) items and \( p(x) \) be the sale price per item if \( x \) items are sold.
The profit \( P(x) \) of selling \( x \) items is calculated as:
\[ P(x) = xp(x) - C(x) \]
where \( xp(x) \) represents the revenue and \( C(x) \) represents the costs.
The average profit per item when \( x \) items are sold is:
\[ \frac{P(x)}{x} \]
The marginal profit, which approximates the profit obtained by selling one more item given that \( x \) items have already been sold, is:
\[ \frac{dP}{dx} \]
Consider the following cost functions \( C \) and price functions \( p \):
\[ C(x) = -0.01x^2 + 130x + 100 \]
\[ p(x) = 300 - 0.1x \]
with \( a = 600 \).
**Part (a) - Finding the Profit Function \( P \):**
To find the profit function \( P \), we use the given formulas:
1. **Revenue Function**: \( R(x) = xp(x) = x(300 - 0.1x) \)
2. **Cost Function**: \( C(x) = -0.01x^2 + 130x + 100 \)
**Steps to Find \( P(x) \):**
\[ P(x) = R(x) - C(x) \]
\[ P(x) = x(300 - 0.1x) - (-0.01x^2 + 130x + 100) \]
\[ P(x) = (300x - 0.1x^2) + 0.01x^2 - 130x - 100 \]
\[ P(x) = 300x - 130x - 100 - 0.09x^2 \]
\[ P(x) = 170x - 100 - 0.09x^2 \]
The profit function \( P(x) \) is:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3b13cfd-0bfc-4fad-ae09-6a6a47693799%2Fb9f3b91b-ee9f-4044-b250-1de2697687e1%2Fl1aqouf_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Profit Functions in Economics**
In this section, we explore how to calculate the profit function \( P(x) \), given cost and price functions. Let's delve into the provided formulas and examples.
**Problem Statement:**
Let \( C(x) \) represent the cost of producing \( x \) items and \( p(x) \) be the sale price per item if \( x \) items are sold.
The profit \( P(x) \) of selling \( x \) items is calculated as:
\[ P(x) = xp(x) - C(x) \]
where \( xp(x) \) represents the revenue and \( C(x) \) represents the costs.
The average profit per item when \( x \) items are sold is:
\[ \frac{P(x)}{x} \]
The marginal profit, which approximates the profit obtained by selling one more item given that \( x \) items have already been sold, is:
\[ \frac{dP}{dx} \]
Consider the following cost functions \( C \) and price functions \( p \):
\[ C(x) = -0.01x^2 + 130x + 100 \]
\[ p(x) = 300 - 0.1x \]
with \( a = 600 \).
**Part (a) - Finding the Profit Function \( P \):**
To find the profit function \( P \), we use the given formulas:
1. **Revenue Function**: \( R(x) = xp(x) = x(300 - 0.1x) \)
2. **Cost Function**: \( C(x) = -0.01x^2 + 130x + 100 \)
**Steps to Find \( P(x) \):**
\[ P(x) = R(x) - C(x) \]
\[ P(x) = x(300 - 0.1x) - (-0.01x^2 + 130x + 100) \]
\[ P(x) = (300x - 0.1x^2) + 0.01x^2 - 130x - 100 \]
\[ P(x) = 300x - 130x - 100 - 0.09x^2 \]
\[ P(x) = 170x - 100 - 0.09x^2 \]
The profit function \( P(x) \) is:
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