# Solve the problem

• June 25th 2008, 12:57 PM
Mr.Huge
Solve the problem
A water rocket is launched upward with an initial velocity of 48ft/sec. Its height h, in feet, after t seconds is given by h = 48t - 16t^2. Determine the height of the rocket 2.5 sec after it has been lanuched.

My work (im lost)

H = 48t - 16t^2

Step 1 (t + 12) (t + 4t)

Step 2 cancel out t left with (-12) (4t)

Step 3 simplify -12/4 4t/4t

Step 4 i am so lost your help is appreciated proper steps please.

Thanks

Mr.H
• June 25th 2008, 01:03 PM
Moo
Hello,

Quote:

Originally Posted by Mr.Huge
A water rocket is launched upward with an initial velocity of 48ft/sec. Its height h, in feet, after t seconds is given by h = 48t - 16t^2. Determine the height of the rocket 2.5 sec after it has been lanuched.

My work (im lost)

H = 48t - 16t^2

Step 1 (t + 12) (t + 4t)

Step 2 cancel out t left with (-12) (4t)

Step 3 simplify -12/4 4t/4t

Step 4 i am so lost your help is appreciated proper steps please.

Thanks

Mr.H

What were you trying to do ? :eek:

$h(t)=48t-16t^2$

And you want the height when $t=2.5$

$h(2.5)=48 \cdot 2.5-16 \cdot (2.5)^2$

..
• June 25th 2008, 01:25 PM
masters
Quote:

Originally Posted by Moo
Hello,

What were you trying to do ? :eek:

$h(t)=48t-18t^2$

And you want the height when $t=2.5$

$h(2.5)=48 \cdot 2.5-18 \cdot (2.5)^2$

..

$\color{red}h(t)=48t-16t^2$

I know....same method! Hard to get a post in these days. I have to take what I can get.http://www.clicksmilies.com/s1106/ak...smiley-069.gif
• June 25th 2008, 01:33 PM
Mr.Huge
im still lost
i still don't get how to work the problem. Can you be more detailed? Give me the run down of how to do this type of problem? I dont get it at all..sorry to sound so slow...this is all new to me.

Mr.H
• June 25th 2008, 01:40 PM
masters
Try these projectile word problems here....Quadratic Word Problems: Projectile Motion

There are several examples.

Good luck!