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A stone is launched vertically upward from a cliff 48 ft above the ground at a speed of 32 ft/s. Its height above the ground t seconds after the launch is given by s = -16t² + 32t + 48 for 0 ≤ t ≤ 3. When does the stone reach its maximum height?
Find the derivative of s.
s' = [ ]
The stone reaches its maximum height at [ ] s. (Simplify your answer.)

Sagot :

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