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

The equation modeling the height of the ball at time,t is given by h=-16t²+48t+6
a. In how many seconds does the ball reach its maximum height?
Time taken for the ball to reach the maximum height will be given by:
h'(t)=-32t+48=0
finding the value of t we get:
32t=48
t=48/32
t=1.5 seconds
Thus the time taken for the ball to reach the maximum time is 1.5 seconds

b] What is the ball’s maximum height?
The maximum height will be:
h(1.5)=-16(1.5)²+48(1.5)+6
h(1.5)=42 fee