A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function
h = –16t2 + 32t + 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? 1 s; 54 ft 2 s; 6 ft 2 s; 22 ft 1 s; 22 ft
For this case we have the following equation: h = -16t2 + 32t + 6 Deriving we have: h '= -32t + 32 Equaling to zero: -32t + 32 = 0 We clear the time: t = 32/32 t = 1 s We are now looking for the maximum height: h (1) = -16 * (1) ^ 2 + 32 * (1) + 6 h (1) = 22 feet Answer: The ball reaches its maximum height in: t = 1 s The ball's maximum height is: h (1) = 22 feet option: 22 ft, 1 s
