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?
Accepted Solution
A:
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]