Posted by **Nisabel** on Friday, March 28, 2014 at 10:09pm.

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

- math -
**Damon**, Friday, March 28, 2014 at 10:37pm
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

- math -
**Bosnian**, Friday, March 28, 2014 at 10:41pm
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

- math -
**Nisabel**, Friday, March 28, 2014 at 10:46pm
Thank you omygosh you guys saved me!

- math -
**Damon**, Friday, March 28, 2014 at 11:16pm
You are welcome:)

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