Q14

A stone is launched vertically upward from a cliff 256 ft above the ground at a speed of 96 ft/s. Its height above the ground t seconds after the launch is given by s = - 161 + 96t + 256 for 0 Sts 8. When does the stone reach its maximum height? Find the derivative of s . s' = 0 The stone reaches its maximum height at s. (Simplify your answer.)