7. A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h=-16t^2+36t+9. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?

A. 1.13 s; 69.75 ft

B. 1.13 s; 29.25 ft

C. 2.25 s; 9 ft

D. 1.13 s; 31.5 ft

The answer is 1.13;29.25

Step-by-step explanation:

The ball reaches its maximum height in 1.13 seconds. b) The ball's maximum height is 30.25 feet.

This sort of problem is solved easily by a graphing calculator.

a) The ball reaches its maximum height in 1.13 seconds.

b) The ball's maximum height is 30.25 feet.

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Since you want to know when and where the peak value of the function occurs, it is convenient to put it into vertex form.

 h = -16(t² -(9/4)t) + 10

 h = -16(t² -(9/4)t +(9/8)²) +10 +16(9/8)²

 h = -16(t -9/8)² + 10 + 81/4

 h = -16(t -9/8)² + 30 1/4

The vertex of the function is (9/8, 30 1/4) ≈ (1.13, 30.25).

h = -16t^2 + 36t + 9

At maximum height h' = 0,

h' = -32t + 36 = 0

32t = 36

t = 36/32 = 1.13 s.

Maximum height = -16(1.125)^2 + 36(1.125) + 9 = 29.25 ft

Answer is option d (1.13 s, 29.25 ft)

The answer is 1.13;29.25

x=6 and x=3

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