Question

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=-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?

Answer 1

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.