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=-16x^2+36t+5. 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?

There are many ways to solve this...however without calculus or derivations from physics, the simplest is to find the midpoint of the two zeros of the function:

(*note, that this will occur for -b/(2a), where a and b are from the general quadratic ax^2+bx+c.  So the time when the ball is at its maximum height is -36/(2*-16),  -36/-32, 1.125 seconds.  And the maximum/minimum height will be (4ac-b^2)/(4a), 25.25 ft.  In general the vertex, minimum/maximum point of a parabola will always be (-b/(2a), (4ac-b^2)/(4a) if you care to commit such to memory.)

So back to the midpoint of the zeros.  Simply use the quadratic formula to find when h=0

t=(-b±√(b^2-4ac))/(2a)

t=(-36±√1616)/-32  regardless of what is under the radical the midpoint for t is simply

t=-36/-32=1.125 seconds

You then can plug this value in for t in the equation h(t) to get the maximum height of 25.25 ft.

31/6= 19/6 19/6 *1/7=19/42