# Thread: Solve the problem

1. ## 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

2. Hello,

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 ?

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

And you want the height when $\displaystyle t=2.5$

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

..

3. Originally Posted by Moo
Hello,

What were you trying to do ?

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

And you want the height when $\displaystyle t=2.5$

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

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

I know....same method! Hard to get a post in these days. I have to take what I can get.

4. ## 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

5. Try these projectile word problems here....Quadratic Word Problems: Projectile Motion

There are several examples.

Good luck!