Question 389256
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Hi,  
Note: the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} 
where(h,k) is the vertex 
  h(t)= ­-16tē + 16t + 480  Believe there should be a minus squared term   
h(t)= ­-16t^2 + 16t + 480  completeing the square to find the vertex
h(t)= ­-16(t^2 - t) + 480
h(t)= -­16[(t - 1/2)^2 - 1/4] + 480
h(t)= ­-16(t - 1/2)^2 + 4 + 480
h(t)= ­-16(t + 1/2)^2 + 484 vertex is Pt(1/2,484) or ordered pair (t,h(t))
a. How long did it take for Jason to reach his maximum height? 1/2  sec
b. What was the highest point that Jason reached? 484ft
c. Jason hit the water after how many seconds?
h(t)= 0 = ­­-16tē + 16t + 480
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{t = (-16 +- sqrt(30976))/(-32) }}}
{{{t = (-16 +- 176)/(-32) }}}
  t = 160/32 = -5 extraneous solution
  t = 192/32 = 6 sec