SOLUTION: /Word problem to solve - very confused Okay here goes!!! It is a launch of a cannon ball Cannon ball should have initial vertical velocity of 15 feet per second. the cannon is

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: /Word problem to solve - very confused Okay here goes!!! It is a launch of a cannon ball Cannon ball should have initial vertical velocity of 15 feet per second. the cannon is      Log On


   



Question 89408: /Word problem to solve - very confused
Okay here goes!!!
It is a launch of a cannon ball
Cannon ball should have initial vertical velocity of 15 feet per second. the cannon is on a hill that is 10 ft high. Keep in mind that the barrel of teh cannon is 4 ft. off the ground (on the hill) and the cannon ball is launched from the top of the cannon.
Okay - I need to use the following model:
h=-16t^+vt + h
h=height
v=initial vertical velocity (ft per second)
h=inital height (ft)
t-time in motion - (seconds)
I have to find a function that models the path of the cannon ball
Graph of the path
Maximum height of the cannon ball above level ground and
how long after the cannon ball is launched does it reach its max height
i can't believe this - HELP!!!!!!!!!!!!!!!

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Cannon ball should have initial vertical velocity of 15 feet per second. the cannon is on a hill that is 10 ft high. Keep in mind that the barrel of teh cannon is 4 ft. off the ground (on the hill) and the cannon ball is launched from the top of the cannon.
Okay - I need to use the following model:
h=-16t^+vt + h
h=height
v=initial vertical velocity (ft per second)
h=inital height (ft)
t-time in motion - (seconds)
I have to find a function that models the path of the cannon ball
h(t) = -16t^2 + 15t + 14
-------------------------
Graph of the path
graph%28400%2C300%2C-5%2C10%2C-5%2C20%2C-16x%5E2%2B15x%2B14%29
-------------------------
Maximum height of the cannon ball above level ground and
how long after the cannon ball is launched does it reach its max height
Max occurs at time t= -b/2a = -15/-32 seconds
H(15/32) = -16(15/32)^2+15(15/32)+14 = 17.516 ft
=============
Cheers,
Stan H.