An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is 50 ft per sec. The heights of the ball in feet is given by the equation s= -2.71² +50t+6.5, where t is the number of seconds after the ball was thrown. Complete parts a and b. a. After how many seconds is the ball 18 ft above the moon's surface? After seconds the ball will be 18 ft above the moon's surface. (Round to the nearest hundredth as needed. Use a comma to separate answers as needed.) ents Libran
An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is 50 ft per sec. The heights of the ball in feet is given by the equation s= -2.71² +50t+6.5, where t is the number of seconds after the ball was thrown. Complete parts a and b. a. After how many seconds is the ball 18 ft above the moon's surface? After seconds the ball will be 18 ft above the moon's surface. (Round to the nearest hundredth as needed. Use a comma to separate answers as needed.) ents Libran
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Problem: Motion of a Baseball on the Moon
An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is 50 ft per sec. The height \( s \) of the ball in feet is given by the equation:
\[ s(t) = -2.7t^2 + 50t + 6.5 \]
where \( t \) is the number of seconds after the ball was thrown. Complete parts a and b.
#### a. After how many seconds is the ball 18 ft above the moon's surface?
After \(\_\_\_\_\_\) seconds the ball will be 18 ft above the moon's surface.
(Round to the nearest hundredth as needed. Use a comma to separate answers as needed.)
---
### Explanation
1. **Equation Explanation:**
The equation provided is a quadratic equation in the form:
\[ s(t) = at^2 + bt + c \]
Here:
- \( a = -2.7 \)
- \( b = 50 \)
- \( c = 6.5 \)
2. **Finding the Time:**
To find the time when the ball reaches a height of 18 feet above the moon's surface, substitute \( s(t) = 18 \) into the equation and solve for \( t \):
\[ 18 = -2.7t^2 + 50t + 6.5 \]
Rearrange this to standard quadratic form:
\[ -2.7t^2 + 50t + 6.5 - 18 = 0 \]
\[ -2.7t^2 + 50t - 11.5 = 0 \]
Then, solve this quadratic equation for \( t \).
A detailed step-by-step process can include using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = -2.7 \)
- \( b = 50 \)
- \( c = -11.5 \)
This will give two values for \( t \), indicating two points in time when the ball reaches 18 feet.
3. **Graphical Representation:**
If there are any graphs or diagrams given in the problem, they would typically plot the height \( s \) of the ball](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fba446a50-8bb7-4794-9fc8-884b5dcd25dd%2F1c05f0d5-418a-4b20-a714-ab878ca9bbde%2F1c2yreb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem: Motion of a Baseball on the Moon
An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is 50 ft per sec. The height \( s \) of the ball in feet is given by the equation:
\[ s(t) = -2.7t^2 + 50t + 6.5 \]
where \( t \) is the number of seconds after the ball was thrown. Complete parts a and b.
#### a. After how many seconds is the ball 18 ft above the moon's surface?
After \(\_\_\_\_\_\) seconds the ball will be 18 ft above the moon's surface.
(Round to the nearest hundredth as needed. Use a comma to separate answers as needed.)
---
### Explanation
1. **Equation Explanation:**
The equation provided is a quadratic equation in the form:
\[ s(t) = at^2 + bt + c \]
Here:
- \( a = -2.7 \)
- \( b = 50 \)
- \( c = 6.5 \)
2. **Finding the Time:**
To find the time when the ball reaches a height of 18 feet above the moon's surface, substitute \( s(t) = 18 \) into the equation and solve for \( t \):
\[ 18 = -2.7t^2 + 50t + 6.5 \]
Rearrange this to standard quadratic form:
\[ -2.7t^2 + 50t + 6.5 - 18 = 0 \]
\[ -2.7t^2 + 50t - 11.5 = 0 \]
Then, solve this quadratic equation for \( t \).
A detailed step-by-step process can include using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = -2.7 \)
- \( b = 50 \)
- \( c = -11.5 \)
This will give two values for \( t \), indicating two points in time when the ball reaches 18 feet.
3. **Graphical Representation:**
If there are any graphs or diagrams given in the problem, they would typically plot the height \( s \) of the ball
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Transcribed Image Text:An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in.
is 50 ft per sec. The heights of the ball in feet is given by the equation s= -2.7
seconds after the ball was thrown. Complete parts a and b.
a. After how many seconds is the ball 12 ft above the moon's surface?
After 10 seconds the ball will be 12 ft above the moon's surface.
(Round to the nearest hundredth as needed. Use a comma to separate answers a:
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