Question

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?

Answer 1

That is the vertex
for y=ax^2+bx+c
the x value of teh vertex is $\frac{-b}{2a}$
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 $\frac{-(48)}{2(-16)}=\frac{48}{32}=\frac{3}{2}$
subsituting it for x we get
$h=-16(\frac{3}{2})^2+48(\frac{3}{2})+6$
$h=42$


it reaches the max height of 42ft at 1.5 seconds

Answer 2

1.5 sec and will reach a height of 42 feet