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

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right