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}=-32 t + 16 ) the stone reaches its maximum height at ( square ) s. (simplify your 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:

\(t = 0.5\)