Question 1187323: An object is thrown upward at a speed of 51 feet per second by a machine from a height of 11 feet off the ground.
The height h of the object after t seconds can be found using the equation
h= −16t^2 + 51t +11
When will the height be 34 feet?
When will the object reach the ground?
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
An object is thrown upward at a speed of 51 feet per second by a machine from a height of 11 feet off the ground.
The height h of the object after t seconds can be found using the equation
h= −16t^2 + 51t +11
When will the height be 34 feet?
When will the object reach the ground?
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To answer first question, solve this quadratic equation
-16t^2 + 51t + 11 = 34.
To answer second question, solve this quadratic equation
-16t^2 + 51t + 11 = 0
and use the greatest (positive) root.
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In this site, there is a bunch of lessons on a projectile thrown/shot/launched vertically up
- Problem on a projectile moving vertically up and down
- Problem on an arrow shot vertically upward
- Problem on a ball thrown vertically up from the top of a tower
- Problem on a toy rocket launched vertically up from a tall platform
Consider these lessons as your textbook, handbook, tutorials and (free of charge) home teacher.
Read them attentively and learn how to solve this type of problems once and for all.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the topic "Projectiles launched/thrown and moving vertically up and dawn".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
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