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. 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

H = -16t • 2 + 48t + 6
h = -32t + 48t + 6
h = 16t + 6

Answer 2

Step-by-step explanation:

It is given that,

Initial velocity of the ball, u = 48 ft/s

The height h in feet after t seconds is given by the function as :

$h=-16t^2+48t+6$........(1)

Maximum height reached by the ball is given by differentiating equation (1) wrt t as :

$\dfrac{dh}{dt}=0$

$\dfrac{d(-16t^2+48t+6)}{dt}=0$

-32 t + 48 = 0

t = 1.5 seconds

Put the value of t in equation (1) as :

$h=-16(1.5)^2+48(1.5)+6$

h = 42 feets

So, the maximum height of the ball is 42 feets . The time taken to reach its maximum height is 1.5 seconds. Hence, this is the required solution.