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?
The equation modeling the height of the ball at time,t is given by h=-16t²+48t+6 <span>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]</span><span> What is the ball’s maximum height? The maximum height will be: h(1.5)=-16(1.5)</span>²+48(1.5)+6 h(1.5)=42 fee
