



When
you throw something into the air,



it goes
up (probably). 


As soon
as it leaves your hand, gravity starts slowing it down. 


At some
point it stops going up and starts coming back down. 


This
equation tells us how high IN FEET the thing is in the air 


at some
time after it is thrown. 





Height
= 16t^{2} + V_{o}t + h_{o}






Now
before you can do anything with this mess 


you need
to know what all of those letters mean. 





t = time
in seconds since you let the thing go 


V_{o}
= the velocity (speed) the
thing was going when it left your hand 


h_{o}
= how high your hand was off
the ground when you let go of the thing 


Height
= how high the thing is at any
time "t." 





In this
problem, we have t instead of X as the independent variable. 


The
Algebra still all works he same way. 


If the t
really bothers you, you can use X instead. 





Example: 


A ball is
thrown upward at a speed of 50 feet per second 


(that's
the V_{o} part) 


It is 6
feet in the air when it leaves the person's hand 


(that's
the h_{o} part) 


How high
above the person's hand does it get??? 





OK, let's
see what numbers we know: 





t = it's
the unknown, it's an independent variable (like X). 


V_{o}
= 50 feet per second 


h_{o}
= 6 feet 


Height
= another unknown. It depends
on t, so it's a dependent variable (like Y) 





So when
we put the numbers we know into the equation we get: 





Height
= 16t^{2} + 50t + 6






We work
this problem exactly the same way as if it was Y = 16X ^{2}
+ 50X + 6 





Since the
number next to the squared variable is negative (16), 


the
parabola opens down. 





So here
we have: 


Height
= 16t^{2} + 50t + 6



So: 


A
= 16 B = 50 C = 6






The
vertex (top) of the parabola happens at: 








About a
second and a half ...






Now we
can use that number to see how high it is at that time. 





Now
here's the tricky part. 


The
question DID NOT ask how high the ball was in the air. 


It asked
how high the ball was ABOVE THE PERSON'S HAND. 


The
person's hand is 6 feet in the air, 


so the
ball is about 39 feet above the person's hand! 





DON'T GET
CAUGHT ON THAT ONE!!!!! 





There is
another thing that you might get asked 


on a
problem like this. 


That is,
how long until the ball hits the ground? 





To solve
that, 


we need
to find the time when the height of the ball equals zero. 


That is,
when it lands. 





So: 


Height
= 16t^{2} + 50t + 6



is: 


0
= 16t^{2} + 50t + 6






Now we
need the quadratic formula ... 








And here 


A
= 16 B = 50 C = 6



So: 





Saying
that a ball hit the ground at .11 seconds 


doesn't
have much meaning here so we can ignore that one. 


Saying
that the ball hit the ground after 3.24 seconds does mean something, 


so that's
our answer. 





So just
where did that .11 seconds come from anyway? 





Here's
where ... 





The shape
of the graph is an upside down parabola. 


That is
also what the flight path of the ball looks like. 











But a
parabola really doesn't start or end anywhere, 


it goes
on forever. 


That
means the parabola will cross the "ground" in two places. 





And
that's where the other answer comes from. 





copyright 2005 Bruce Kirkpatrick 
