Question 89408
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
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Graph of the path
{{{graph(400,300,-5,10,-5,20,-16x^2+15x+14)}}}
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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
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Cheers,
Stan H.