A ball is thrown into the air with an upward velocity of 48 ft/s. Its height h in feet after t seconds is given by the function h = –16t2 + 48t + 6. 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?

that is the vertex
for y=ax^2+bx+c
the x value of teh vertex is [tex]\frac{-b}{2a}[/tex]
the y value is found by subsituting the x value of the vertex for x
so
h=-16t^2+48t+6
a=-16
b=48
c=6

x value of vertex is [tex]\frac{-(48)}{2(-16)}=\frac{48}{32}=\frac{3}{2}[/tex]
subsituting it for x we get
[tex]h=-16(\frac{3}{2})^2+48(\frac{3}{2})+6[/tex]
[tex]h=42[/tex]


it reaches the max height of 42ft at 1.5 seconds