# Thread: College Algebra Urgent Help!!

1. ## College 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:

Thanks

2. Good luck.

-Andy

3. Originally Posted by bulldogben21
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:

Thanks
Abender: I'm a little slow today trying to make it look pretty.

4. Man thank you so much. You have no idea how much that helped

5. 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.

6. Oooops, my apologies for introducing calculus on a thread titled "College Algebra Urgent Help!!" Thank you for catching this masters.