A ball is thrown upward from the top of a 100-foot building with an initial velocity of 32 feet per second. The height of the ball above the ground after t seconds is given by the function h(t) = 100 + 32t - 16t². What is the instantaneous rate of change of the ball's height when t = 2 seconds?

Core Concept

To find the instantaneous rate of change, we need to find the derivative of h(t). h'(t) = 32 - 32t. At t = 2, h'(2) = 32 - 32(2) = 32 - 64 = 0 ft/s. This means the ball is at its maximum height at t = 2 seconds and is momentarily at rest before beginning its descent. Option B is incorrect because it's the initial velocity, not the velocity at t = 2. Option C is incorrect because it's half the correct value. Option D is incorrect because it's the initial velocity, not the velocity at t = 2.