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 = - 16t2 + 36t + 9. 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? 1.13 s; 29.25 ft 1.13 s; 31.5 ft 2.25 s; 9 ft 1.13 s; 69.75 ft

approximately 1.13 seconds is when the max height is obtained

29.25 ft is the max height

Step-by-step explanation:

Maximum/minimum you should automatically go to vertex if you are dealing with a parabola or a quadratic; I'm talking about something in this form y=ax^2+bx+c.

The x-coordinate of the vertex can be found by computing - b / (2a)

Or in this case the t-coordinate.

a=-16

b=36

c=9

Plug in (you don't need c for this) - 36 / (2*-16) = -36/-32 (reduce) = 9/8

(divide; put in calc 9 divided by 8) = 1.125

t represented the seconds so we done with part A which is 1.125 seconds

Now for B, all you have to do once you found the x - (or t - in this case) coordinate, plug it into your equation that relates x (or t in this case) and y (or h in this case).

h=-16 (1.125) ^2+36 (1.125) + 9

I'm just going to put - 16 (1.125) ^2+36 (1.125) + 9 into my calculator exactly as it appears which is 29.25 ft.