SOLUTION: A flare is shot upward from the edge of a 88 foot tall cliff with an initial velocity of 82 ft/sec. What is the maximum height above the ground this flare will go in feet? Rou

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Question 1115426: A flare is shot upward from the edge of a 88 foot tall cliff with an initial velocity of 82 ft/sec. What is the maximum height above the ground this flare will go in feet?
Round your answer to 1 decimal place.

Found 2 solutions by KMST, ikleyn:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
This is a classical physics problem, and classical physics should be intuitive, because we see it in action in the world around us.
Alternatively, we can memorize "formulas" for every "type" of problem without really understanding.

THE SIMPLE/FIFTH-GRADER SOLUTION:
Somewhere in the physics book, it is stated that the acceleration of gravity increases the downward speed of an object by 32 feet per second every second:
g=32ft%2Fs%5E2 (with s as the symbol for second).
So, it will take 82%2F32=2.5625 seconds for gravity to get that initial upwards velocity down to zero, before the flare remain star having increasing downwards velocity.
During those initial 2.5625 seconds the upwards velocity decreased steadily/linearly from %2282+ft+%2F+s%22 to %220+ft+%2F+s%22 .
.
During that time the average upwards velocity (in ft/s) was
%2882%2B0%29%2F2=41 ,
and the upwards distance covered (in feet) was
41%2A2.5625=105.0625 .
We can round that to 1 decimal place to get 105.1ft .
As the flare started from 88ft above ground,
the maximum height above the ground this flare will go (in feet) is
88%2B105.1=highlight%28193.1%29 .

WITH FORMULAS (maybe your teacher wants that):

The basic formulas for an object moving in one direction with an "initial velocity" v%5B0%5D and a constant acceleration a are
v=v%5B0%5D%2Bat and
d=v%5B0%5Dt%2B%281%2F2%29at%5E2 or v%5Bavg%5D=%28v%2Bv%5B0%5D%29%2F2 (pick one).
In those formulas d is distance covered,
t is time since a time, t=0 ,
that we can conveniently say it is the start of the observation,
an "initial velocity" is the velocity at the start of the observation,
when the velocity was v%5B0%5D .
From those two formulas, and your algebra skills,
you can get any other formula you will be given.

In particular, for projectiles shot upwards, from somewhere om planet Earth,
the height h above the ground is
h=h%5B0%5D%2Bv%7B0%5Dt-16t%5E2 ,
where h%5B0%5D is the initial height above the ground, and t is the time since "launch".
plugging into that formula the numbers from the problem,
h=88%2B82t-16t%5E2 .
In algebra class, you would probably write it as
h=-16t%5E2%2B82t%2B88 ,
because we like to order polynomial terms from highest to lowest exponent.
In algebra 2 courses is the US, you are likely to be told that
a quadratic function f%28x%29=ax%5E2%2Bbx%2Bc has a maximum (if a%3C0 )
at x=%28-b%29%2F%222+a%22 . If a%3E0 it would be a minimum)
(another formula, but you do not need to memorize it if your algebra skills include "completing the square").
h=-16t%5E2%2B82t%2B88 is a quadratic function of t (instead of x ),with a=-16 and b=82 ,
so it has a maximum at t=%28-b%29%2F%222+a%22=%28-82%29%2F%282%2A%28-16%29%29=%28-82%29%2F%28-32%29=5.5625 ,
and for that value of t , in seconds, the height h , in feet, is
h=-16%2A2.5625%5E2%2B82%2A2.5625%2B88=-106.0625%2B210.125%2B88=193.0625 .
Rounding to 1 decimal place, highlight%28h=193.1%29 .

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
A flare is shot upward from the edge of a 88 foot tall cliff with an initial velocity of 82 ft/sec.
What is the maximum height above the ground this flare will go in feet?
Round your answer to 1 decimal place.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The formula for the height above the cliff base as the function of time is

h(t) = -16t^2 + 82t + 88.     (1)


The height is maximum when the quadratic function (1) is maximum.

It happens when  t = -b%2F%282a%29 = -82%2F%282%2A%28-16%29%29 = 82%2F32 = 2.56 seconds.


At this time,  h(t) = -16*2.56^2 + 82*2.56 + 88 = 193.1 ft.


Answer.  The maximum height above the cliff base is 193.1 ft.

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On finding the maximum/minimum of a quadratic function see the 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.

To see other similar solved problems on a projectile thrown/shot/launched vertically up,  look into the lessons
    - 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

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I wrote my text to fix an error in the @KMST calculation of the time to get the maximum height.