Question 1135232: A rock is thrown upward with a velocity of 22 meters per second from the top of a 23 meter high cliff, and it misses the cliff on the way back down. When will the rock be 10 meters from the water, below? Round your answer to two decimal places.
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
The general equation for the height of the rock above the water surface (in meters) is
h(t) = -5*t^2 + 22t + 23
To answer the question, you must determine at what time "t" the height h(t) will be 10, 10 meters above the water surface.
So, you need to solve this equation
-5t^2 + 22t + 23 = 10.
Simplify
5t^2 - 22t - 13 = 0.
Solve this quadratic equation using quadratic formula and keep its positive root.
Do not forget to post your deepest "THANKS" to me for my lesson and my explanations.
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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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