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 + ( 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!

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To find the time it takes for the ball to reach its maximum height, you need to determine the value of "t" where the equation h(t) = -16t^2 + 20t + 2 reaches its maximum point.

In this case, the height function is a quadratic equation in the form h(t) = at^2 + bt + c, where a = -16, b = 20, and c = 2. The maximum point of a quadratic function occurs at the vertex, which is found using the formula t = -b / (2a).

Plugging in the values, we have t = -20 / (2*(-16)). Simplifying this, we get t = -20 / -32 = 5/8 = 0.625.

Rounding to the nearest hundredth, the ball takes approximately 0.63 seconds to reach its maximum height.

To find the maximum height, you need to substitute the value of "t" into the height function h(t).

Plugging in t = 0.63, we have h(0.63) = -16*(0.63)^2 + 20*(0.63) + 2.

Calculating this, h(0.63) ≈ -6.45 + 12.6 + 2 ≈ 8.15.

Rounding to the nearest hundredth, the ball reaches its maximum height of approximately 8.15 feet.