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

The time it takes by the ball to reach maximum height and the maximum height is required.

Time taken to reach maximum height is 1.5 s.

Maximum height the ball reached is 42 ft.

The given function is

$h=-16t^2+48t+6$

Differentiating with respect to time

$h'=-32t+48$

Equating to zero

$0=-32t+48\\\Rightarrow t=\dfrac{48}{32}=1.5$

Double derivative of h

$h''=-32$

As it is negative height will be maximum at 1.5 s.

$h=-16(1.5)^2+48(1.5)+6\\\Rightarrow h=42\ \text{ft}$

Maximum height the ball reached is 42 ft.

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