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? 2.25 s; 10 ft 1.13 s; 30.25 ft 1.13 s; 70.75 ft 1.13 s; 32.5 ft

We can do this by converting the equation to vertex form:-

h = -16t^2 + 36t + 10

= -16(t^2 - 2.25t) + 10

= -16 [ (t  - 1.125)^2 - (1.125)^2] + 10

= -16(t - 1.125)^2 + 20.25 + 10

= -16(t - 1.125)^2 + 30.25  

So the answer is 1.13s and 30.25 ft.

=  (

the height of the surface increases from the center out to the sides of the road.

step-by-step explanation:

as we move out we move away from the x-axis.

and as mentioned in the problem statement that height of a road increases when we move away from the center x.

hence proved that 3rd option

the height of the surface increases from the center out to the sides of the road.

is the correct option.