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
Try these projectile word problems here....Quadratic Word Problems: Projectile Motion
There are several examples.
Good luck!