1. **State the problem:** A ball is thrown from a height of 64 feet. Its height above the ground after $t$ seconds is given by the function $$s(t) = -16t^2 + 64.$$ We need to find the instantaneous velocity of the ball when it hits the ground. 2. **Formula and rules:** The instantaneous velocity is the derivative of the position function with respect to time, i.e., $$v(t) = s'(t).$$ The ball hits the ground when its height is zero, so we first find $t$ such that $$s(t) = 0.$$ Then we evaluate $v(t)$ at that time. 3. **Find when the ball hits the ground:** $$-16t^2 + 64 = 0$$ Divide both sides by $-16$: $$\cancel{-16}t^2 + \cancel{64} = 0 \Rightarrow t^2 - 4 = 0$$ 4. **Solve for $t$:** $$t^2 = 4$$ $$t = \pm 2$$ Since time cannot be negative, $$t = 2$$ seconds. 5. **Find the velocity function:** $$v(t) = \frac{d}{dt}(-16t^2 + 64) = -32t$$ 6. **Calculate instantaneous velocity at $t=2$:** $$v(2) = -32 \times 2 = -64$$ feet per second. 7. **Interpretation:** The negative sign indicates the ball is moving downward when it hits the ground. **Final answer:** The instantaneous velocity of the ball when it hits the ground is $$\boxed{-64}$$ feet per second.