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) = -16t^2 + 32t + 100. What is the instantaneous rate of change of the height of the ball at t = 2 seconds?

Core Concept

To find the instantaneous rate of change, we need to find the derivative of h(t) and evaluate it at t = 2. h'(t) = -32t + 32. At t = 2, h'(2) = -32(2) + 32 = -64 + 32 = -32 ft/s. Option A is the correct answer. Option B is incorrect because it's the velocity at the maximum height, not at t = 2. Option C is the initial velocity, not the velocity at t = 2. Option D is the acceleration multiplied by time, which is not relevant here.