Aball 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 = –16t2 + 36t + 10. a. in how many seconds does the ball reach its maximum height? round to the nearest hundredth if necessary. b. what is the ball’s maximum height?

a. 1.13 s; 30.25 ft

b. 2.25 s; 10 ft

c. 1.13 s; 70.75 ft

d. 1.13 s; 32.5 ft

  a.  1.13 s

  b.  30.25 ft

Step-by-step explanation:

Both questions can be answered by putting the equation into vertex form.

  h = -16(t^2 -9/4t) +10 . . . . factor the leading coefficient from the t terms

Complete the square by adding the square of half the t coefficient inside parentheses and the opposite of the same amount outside parentheses.

  h = -16(t^2 -9/4t +81/64) +10 +81/4

  h = -16(t -1.125)^2 + 30.25

This tells us the vertex of the height curve is at (t, h) = (1.125, 30.25), which rounds to ...

the answer is d

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