Question

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 )

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:

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