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 - 16t^2. What is the instantaneous rate of change of the height of the ball when 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. The derivative h'(t) = 32 - 32t. Evaluating at t = 2, we get h'(2) = 32 - 32(2) = 32 - 64 = -32 ft/s. However, this is the rate of change of height with respect to time, which is the velocity. The question asks for the instantaneous rate of change of height, which is the derivative of h(t). The correct answer is A) 0 ft/s, which represents the velocity at the highest point of the ball's trajectory.