SOLUTION: A ball is thrown vertically upwards from the top of a school building. It’s height, h metres, above the ground, can be modelled by h=18+6t-4t^2, where t is the time in seconds afte

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A ball is thrown vertically upwards from the top of a school building. It’s height, h metres, above the ground, can be modelled by h=18+6t-4t^2, where t is the time in seconds afte      Log On


   



Question 1113068: A ball is thrown vertically upwards from the top of a school building. It’s height, h metres, above the ground, can be modelled by h=18+6t-4t^2, where t is the time in seconds after it leaves the top of the building.
i) how tall is the school building?
ii) at what time will the ball strike the ground?

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
The correct equation written for the Earth conditions, must have coefficient EITHER -4.9 at t^2,  OR -5,


depending on which value of the gravity acceleration,  9.81 m/s^2   or simply rounded 10 m/s^2 is used.


Simply speaking,  your equation is written incorrectly,


so further discussion makes no sense.

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

    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 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.

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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

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.