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 + 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.

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