An arrow is shot into the air so that its horizontal velocity is 37.0 ft/sec and it vertical velocity is 11.0 ft/sec respectively. Find the velocity of the arrow. Round your answer to the nearest ft/sec. 39 feet per second 27 feet per second 35 feet per second 24 feet per second
An arrow is shot into the air so that its horizontal velocity is 37.0 ft/sec and it vertical velocity is 11.0 ft/sec respectively. Find the velocity of the arrow. Round your answer to the nearest ft/sec. 39 feet per second 27 feet per second 35 feet per second 24 feet per second
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
**Problem Statement:**
An arrow is shot into the air so that its horizontal velocity is 37.0 ft/sec and its vertical velocity is 11.0 ft/sec respectively. Find the velocity of the arrow. Round your answer to the nearest ft/sec.
### Multiple Choice Answers:
- 39 feet per second
- 27 feet per second
- 35 feet per second
- 24 feet per second
### Solution Explanation:
To find the velocity of the arrow, we need to calculate the resultant velocity using the Pythagorean theorem. Given:
- Horizontal velocity (\(V_x\)) = 37.0 ft/sec
- Vertical velocity (\(V_y\)) = 11.0 ft/sec
The resultant velocity (\(V_r\)) can be found using the formula:
\[ V_r = \sqrt{V_x^2 + V_y^2} \]
Plugging in the values:
\[ V_r = \sqrt{(37.0)^2 + (11.0)^2} \]
\[ V_r = \sqrt{1369 + 121} \]
\[ V_r = \sqrt{1490} \]
\[ V_r \approx 38.6 \]
Rounding to the nearest ft/sec, the resultant velocity is:
\[ V_r \approx 39 \text{ feet per second} \]
Thus, the correct answer is **39 feet per second**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdc5bcc65-7283-4529-b9b7-ffbb90a5afc1%2Fc7785713-8316-47db-95aa-a6450a950c53%2Fhjkngrc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Projectile Motion Problem
**Problem Statement:**
An arrow is shot into the air so that its horizontal velocity is 37.0 ft/sec and its vertical velocity is 11.0 ft/sec respectively. Find the velocity of the arrow. Round your answer to the nearest ft/sec.
### Multiple Choice Answers:
- 39 feet per second
- 27 feet per second
- 35 feet per second
- 24 feet per second
### Solution Explanation:
To find the velocity of the arrow, we need to calculate the resultant velocity using the Pythagorean theorem. Given:
- Horizontal velocity (\(V_x\)) = 37.0 ft/sec
- Vertical velocity (\(V_y\)) = 11.0 ft/sec
The resultant velocity (\(V_r\)) can be found using the formula:
\[ V_r = \sqrt{V_x^2 + V_y^2} \]
Plugging in the values:
\[ V_r = \sqrt{(37.0)^2 + (11.0)^2} \]
\[ V_r = \sqrt{1369 + 121} \]
\[ V_r = \sqrt{1490} \]
\[ V_r \approx 38.6 \]
Rounding to the nearest ft/sec, the resultant velocity is:
\[ V_r \approx 39 \text{ feet per second} \]
Thus, the correct answer is **39 feet per second**.
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