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}=square )

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}=square )

Answer

Explanation:

Step1: Apply the power rule

The power rule states that if (y = ax^{n}), then (y^\prime=anx^{n - 1}). For (s=-16t^{2}+16t + 192), the derivative of (-16t^{2}) is (-16\times2t^{2-1}=-32t), the derivative of (16t) is (16\times1t^{1 - 1}=16), and the derivative of the constant (192) is (0).

Step2: Combine the derivatives

Using the sum rule of derivatives ((u + v+w)^\prime=u^\prime + v^\prime+w^\prime) (where (u=-16t^{2}), (v = 16t), (w = 192)), we get (s^\prime=-32t + 16).

Answer:

(s^\prime=-32t + 16)