SOLUTION: A rock is thrown upward with a velocity of 18 meters per second from the top of a 36 meter high cliff, and it misses the cliff on the way back down. When will the rock be 4 meters

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Question 1123163: A rock is thrown upward with a velocity of 18 meters per second from the top of a 36 meter high cliff, and it misses the cliff on the way back down. When will the rock be 4 meters from the water, below? Round your answer to two decimal places.
Found 3 solutions by Boreal, Alan3354, ikleyn:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
draw this
h=-9.8t^2+18t+36
want h=4
therefore
-9.8t^2+18t+32=0
t=(-1/19.6)-18+/-sqrt(324+1254.4); sqrt term is equal to 39.73, use negative root only
t=(-1/19.6)(-57.73)=2.95 seconds


Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
A rock is thrown upward with a velocity of 18 meters per second from the top of a 36 meter high cliff, and it misses the cliff on the way back down. When will the rock be 4 meters from the water, below? Round your answer to two decimal places.
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You didn't supply the equation for the trajectory.
Another tutor used -9.8t^2 as the squared term, but it's -4.9t^2 (on Earth).

Answer by ikleyn(52777)   (Show Source): You can put this solution on YOUR website!
.

My experience at this forum says me that the school students usually are very poorly educated regarding this "height" equation:
as a rule, they don't know how this equation was introduced and what its terms mean.

Therefore, I developed some introductory explanations for such students and keep them in my archive.
When it is needed, I release these explanations to the outside:


Let me explain you EVERYTHING about these problems,
from the very beginning to as far as you need to know it NOW.

1.  You may often meet these problems on a projectile thrown vertically upward.

    The equation for the height over the ground usually has ONE OF TWO POSSIBLE forms:

        a)  h(t) = -16*t^2 + v*t + c

                 In this form, the equation is written for the height h(t) over the ground measured in feet.
                 The value of "16" is the half of the value of the gravity acceleration g = 32 ft/s^2.

                 The sign "-" at the first term means that the gravity acceleration is directed down, 
                 while the "y"-axis of the coordinate system is directed vertically up, in the opposite direction.

                 The value of "v" in this equation is the value of the initial vertical velocity. 

                 The value of "c" is the initial height over the ground.

                 The ground level is assumed to be 0 (zero, ZERO). In other words, the origin of the coordinate system is at the ground.


        b)  h(t) = -5*t^2 + v*t + c

                 It is another form of the "height" equation for the same process.

                 In this form, the height h(t) is measured in meters (instead of feet).

                 The value of "5" at the first term is the same half of gravity acceleration, but this time expressed in "m/s^2" units" g = 10 m/s^2.

                       Actually, more precise value is g = 9.8 m/s^2, therefore, sometimes, this equation goes with the first term -4.9.

                 The value of "v" is the vertical velocity, expressed in m/s.

                 The value of "c" is the initial height over the ground in meters.



2.  In any case, when such problems comes from Algebra (as Algebra problems), they are treated in THIS WAY:

    The question "find the maximal height" is the same as "find the maximum of the quadratic form h(t) = -16t^2 + vt + c.

    It doesn't matter that the quadratic function presented as the function of "t" instead of more usual "x" variable.


    Next, when the question is about the maximum/minimum of a quadratic form 

    q(x) = ax^2 + bx + c,

    then Algebra teaches us that the maximum is achieved at   x = .

    
3.  Same problems may come from CALCULUS.  In Calculus, they are treated in this way:

    to find the maximum (minimum), take the derivative and equate it to zero. 

    It will give the equation to find "t".


4.  Same problems may come from PHYSICS.  In Physics, they are treated in this way:

    the maximum height is achieved when the verical velocity becomes equal to zero. 

    It will give the equation to find "t":   t = .


5.  The amazing fact is that different approaches from different branches of Math and Science give the same answer.

    AMAZING ? - Yes, of cource, without doubts for young students.

    AMAZING ? - Yes, but not so much for more mature students, who understand that all these branches of knowledge 

                study the same Nature's phenomenos.  So, the results should (and must) be identical.


---------------
On finding maximum/minimum of a quadratic function see my lessons
    - HOW TO complete the square to find the minimum/maximum of a quadratic function
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola
in this site.

My lessons on a projectile thrown/shot/launched vertically up are
    - 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
in this site.


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  "Finding minimum/maximum of quadratic functions"
and 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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