Height of a ball Word Problem!

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January 23rd 2010, 04:50 PM
Annb

Height of a ball Word Problem!

Hello everyone,

I'm stuck on this problem and would love an explanation:

If a ball is thrown into the air at 64 feet per second from the top of a 100 foot tall building, its height can be modeled by the function
S=100+64t-16t(squared), where S is in feet and t is in seconds.

Find the max height the ball will reach??

Thanks(Rock)
January 23rd 2010, 05:05 PM
skeeter

Quote:

Originally Posted by Annb

If a ball is thrown into the air at 64 feet per second from the top of a 100 foot tall building, its height can be modeled by the function
S=100+64t-16t(squared), where S is in feet and t is in seconds.

Find the max height the ball will reach??

the graph formed by $s(t) = 100+64t-16t^2$ is an inverted parabola ... the ball reaches its maximum height at the vertex of that parabola.

does the expression $\frac{-b}{2a}$ ring a bell ?
January 23rd 2010, 05:07 PM
VonNemo19

Quote:

Originally Posted by Annb

Hello everyone,

I'm stuck on this problem and would love an explanation:

If a ball is thrown into the air at 64 feet per second from the top of a 100 foot tall building, its height can be modeled by the function
S=100+64t-16t(squared), where S is in feet and t is in seconds.

Find the max height the ball will reach??

Thanks(Rock)

Well, this - $s(t)=100+64t-16t^2$ - is simply a parabola that opens downward. Do you know how to find the vertex?
January 23rd 2010, 05:08 PM
Annb

no : (
January 23rd 2010, 05:14 PM
VonNemo19

Quote:

Originally Posted by Annb

no : (

You may need a refresher. Check out these links:

Vertex of a Parabola

The Vertex of a Parabola
January 23rd 2010, 05:14 PM
Annb

is it y=ax squared + bx + c?
January 23rd 2010, 05:22 PM
VonNemo19

Quote:

Originally Posted by Annb

is it y=ax squared + bx + c?

Yes, that is the standard equation of a parabola.

So, in your case...what are the values $a,b,$ and $c$ ?
January 23rd 2010, 05:26 PM
Annb

a=100
b=64
c=16
January 23rd 2010, 05:32 PM
VonNemo19

Quote:

Originally Posted by Annb

a=100
b=64
c=16

No. $a$ is always the number beside the squared variable. $b$ is always the number beside the first degree variable, and $c$ is always the constant term.

So

$a=-16$
$b=64$
$c=100$ .

Do you understand?

Now, use the formula for finding the vertex. Do you know it?
January 23rd 2010, 05:42 PM
Annb

is it -b/2a?
January 23rd 2010, 05:46 PM
VonNemo19

Quote:

Originally Posted by Annb

is it -b/2a?

Very good. But, that's not the whole story. This will give you the time at which the object will reach its maximum height. So...what do you have to do once you have this time?
January 23rd 2010, 05:56 PM
Annb

Ok, so I when I did the equation I got the answer -2. I just apply it to the equation and substitute t for -2 and i got 164!! The answer..

Thank you
January 23rd 2010, 06:03 PM
VonNemo19

Quote:

Originally Posted by Annb

Ok, so I when I did the equation I got the answer -2. I just apply it to the equation and substitute t for -2 and i got 164!! The answer..

Thank you

Well... almost right. Let me ask you a question: Can time ever be negative? The answer is no (at least in this context).

$\frac{-b}{2a}=\frac{-(64)}{2(-16)}=\color{red}{+}2$ .

Now, plugging back in we have

$s(2)=100+64(2)-16(2)^2$

$=100+128-64$

$=164$ .

Sorry for being picky, but precision is key in mathematics.

(Hi)