math
posted by Nisabel .
A ball is thrown into the air with an upward velocity of 20 ft/s. Its height (h) in feet after t seconds is given by the function h(t) = –16t^2 + 20t + 2. How long does it take the ball to reach its maximum height? What is the ball’s maximum height? Round to the nearest hundredth, if necessary
reaches maximum height of 8.25 feet after 1.25 seconds.
reaches a maximum height of 8.25 feet after 0.63 seconds
reaches a maximum height of 0.16 feet after 1.34 seconds
reaches a maximum height of 0.32 feet after 1.34 seconds

assume no calculus allowed
thus complete he square to find the vertex of this parabola
16 t^2  20 t  2 = h
16 t^2 20 t =  h + 2
t^2  5/4 t =  h/16 + 1/8
t^2 5/4 t+ 25/64 = h/16 + 8/64 + 25/64
(t  5/8)^2 = (1/16)(h  33/4)
so in 5/8 of a second it reaches the vertex at 33/4 = 8 1/4 ft 
h ( t ) = a t ^ 2 + b t + c
h ( t ) = – 16 t ^ 2 + 20 t + 2
a =  16
b = 20
c = 2
t  coordinate of minimum point
t =  b / 2 a
t =  20 / [ 2 *  16 ) ]
t =  20 /  32 = 20 / 32 = 4 * 5 / ( 4 * 8 ) = 5 / 8 = 0.625
h  coordinate of minimum point
h =  ( b ^ 2  4 a c ) / 4 a
h =  [ 20 ^ 2  4 * (  16 ) * 2 ] / [ 4 * (  16 ) ]
h =  ( 400 + 128 ) /  64 =  528 /  64 = 528 / 64 = 16 * 33 / ( 16 * 4 ) = 33 / 4 = 8.25
Round to the nearest hundredth:
t = 0.63 , h = 8.25 
Thank you omygosh you guys saved me!

You are welcome:)
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