SOLUTION: A person standing close to the edge on top of a 96-foot building throws a ball vertically upward. The quadratic function h(t)=−16t^2+80t+96 models the ball's height about the gr

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Question 1154647: A person standing close to the edge on top of a 96-foot building throws a ball vertically upward. The quadratic function h(t)=−16t^2+80t+96
models the ball's height about the ground, h(t), in feet, t seconds after it was thrown.
What is the maximum height of the ball?
How many seconds does it take until the ball hits the ground?
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

What is the maximum height of the ball?
h%28t%29=-16t%5E2%2B80t%2B96 -> it is a parabola that opens downward, max is at vertex; so, write equation in vertex form

+h%28t%29=%28-16t%5E2%2B80t%29%2B96....complete square
h%28t%29=-16%28t%5E2%2B5t%29%2B96
h%28t%29=-16%28t%5E2%2B5t%2Bb%5E2%29-%28-16b%5E2%29%2B96-> b=5%2F2
h%28t%29=-16%28t%5E2%2B5t%2B%285%2F2%29%5E2%29%2B16%285%2F2%29%5E2%2B96
h%28t%29=-16%28t%2B5%2F2%29%5E2%2B16%2825%2F4%29%2B96
h%28t%29=-16%28t%2B5%2F2%29%5E2%2B4%2825%29%2B96
+h%28t%29=-16%28t%2B5%2F2%29%5E2%2B100%2B96
+h%28t%29=-16%28t%2B5%2F2%29%5E2%2B196
h=5%2F2, k=196+=> vertex is at (5%2F2,196)

How many seconds does it take until the ball hits the ground?

-16t%5E2%2B80t%2B96=0

t=%28-80%2B-sqrt%2880%5E2-4%28-16%29%2A96%29%29%2F%282%28-16%29%29

t=%28-80%2B-sqrt%286400%2B6144%29%29%2F-32

t=%28-80%2B-sqrt%2812544%29%29%2F-32

t=%28-80%2B-112%29%2F-32


t=%28-80%2B112%29%2F-32=>t=-1-> disregard negative solution
t=%28-80-112%29%2F-32=>t=6
=> it takes 6 seconds until the ball hits the ground

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
A person standing close to the edge on top of a 96-foot building throws a ball vertically upward. The quadratic function h(t)=−16t^2+80t+96
models the ball's height about the ground, h(t), in feet, t seconds after it was thrown.
What is the maximum height of the ball?
How many seconds does it take until the ball hits the ground?
You don't have to do what the other person did. It's TOTALLY UNNECESSARY!

matrix%281%2C3%2C+h%28t%29%2C+%22=%22%2C+-+16t%5E2+%2B+80t+%2B+96%29

Maximum height occurs at: 

With t = 2.5, maximum height of ball, or:  

To find the time it takes for the ball to hit the ground, we set h(t) = 0. We then get: 

0 = (t - 6)(t + 1)

0 = t - 6                OR                 0 = t + 1

Time it takes to hit the ground, or: highlight_green%28matrix%281%2C4%2C+t%2C+%22=%22%2C+6%2C+seconds%29%29        OR            t = - 1 (ignore)

In other words, you don't have to COMPLETE the SQUARE, and you certainly DON'T have to use the quadratic equation formula! Unless you want to!!