Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t² + 32t + 48.
Determine the velocity v of the stone after t seconds.

Verified step by step guidance

1. Step 1: Understand that the velocity of the stone is the derivative of the height function s(t) with respect to time t. This is because velocity is the rate of change of position with respect to time.
2. Step 2: Write down the height function s(t) = -16t^2 + 32t + 48.
3. Step 3: Differentiate the height function s(t) with respect to t to find the velocity function v(t). Use the power rule for differentiation: if f(t) = at^n, then f'(t) = n*at^(n-1).
4. Step 4: Apply the power rule to each term in s(t): the derivative of -16t^2 is -32t, the derivative of 32t is 32, and the derivative of 48 is 0.
5. Step 5: Combine the derivatives to write the velocity function: v(t) = -32t + 32.