a stone is launched vertically upward from a cliff 192 ft above the ground at a speed of 16 ft/s. its height…

a stone is launched vertically upward from a cliff 192 ft above the ground at a speed of 16 ft/s. its height above the ground t seconds after the launch is given by ( s=-16 t^{2}+16 t + 192 ) for ( 0 leq t leq 4 ). when does the stone reach its maximum height? find the derivative of s. ( s^{prime}=-32 t + 16 ) the stone reaches its maximum height at ( square ) s. (simplify your answer)

a stone is launched vertically upward from a cliff 192 ft above the ground at a speed of 16 ft/s. its height above the ground t seconds after the launch is given by ( s=-16 t^{2}+16 t + 192 ) for ( 0 leq t leq 4 ). when does the stone reach its maximum height? find the derivative of s. ( s^{prime}=-32 t + 16 ) the stone reaches its maximum height at ( square ) s. (simplify your answer)

Answer

Explanation:

Step1: Set the derivative equal to zero

The derivative (s'=-32t + 16). At the maximum - height, the velocity (derivative of the position function) is zero. So we set (s'=0), which gives the equation (-32t+16 = 0).

Step2: Solve the equation for (t)

Add (32t) to both sides of the equation (-32t + 16=0): (16=32t). Then divide both sides by 32: (t=\frac{16}{32}).

Answer:

(t = 0.5)