Hey everyone. I need some major help. Here is my question.

The height of a ball thrown up at 320 feet per second is given by:

h(t)=-16tsquared +320t+5. Find the maximum height the ball reaches. Please help!

Thanks

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- July 27th 2008, 02:05 PMbulldogben21College Algebra Urgent Help!!
Hey everyone. I need some major help. Here is my question.

The height of a ball thrown up at 320 feet per second is given by:

h(t)=-16tsquared +320t+5. Find the maximum height the ball reaches. Please help!

Thanks - July 27th 2008, 02:27 PMabender
Good luck.

-Andy - July 27th 2008, 02:32 PMmasters
Abender: I'm a little slow today trying to make it look pretty.

http://www.texify.com/img/%5CLARGE%5...C%20height.gif - July 27th 2008, 02:33 PMbulldogben21
Man thank you so much. You have no idea how much that helped

- July 27th 2008, 02:33 PMMatt Westwood
The ball goes up, reaches its maximum height,

**stops**, then starts coming down again.

The operative word there is "stops".

So if you can find the value of h when t is equal to zero you're made.

From the equations of motion you've got

h = u t + 1/2 a t^2

where u = 320, a = -32 and h is the value you're trying to find. This matches the equation you've been given, except for the 5. So write it:

h - 5 = -16 t^2 + 320 t

The 5 signifies the extra 5 feet above the ground you start throwing the ball upwards from.

So now you can look at the equation of motion that goes:

v^2 = u^2 + 2 a (h-5)

You've got v, it's equal to zero (see above), and h is what you want to find.

So you have h-5 = (v^2 - u^2) / (2 a)

and plugging in the numbers you get

h-5 = (0 - 320^2) / (2 x (-32))

that should be enough, you should be able to work it out from there. - July 27th 2008, 02:36 PMabender
Oooops, my apologies for introducing calculus on a thread titled

**"College Algebra**Urgent Help!!" Thank you for catching this masters.