A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 36t + 10. 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?
[tex]h(t)=-16t^{2}+36t+10[/tex] [tex]h'(t)=-32t+36[/tex] When h'(t) = 0 the maximum height is reached (turning point). 32t = 36 t = 1.13s The maximum height is found by getting the value of h(1.13) as follows:[tex]h(1.13)=-16(1.13)^{2}+(36\times1.13)+10=30.25 feet.[/tex]
