1. **State the problem:** A ball rolls at an initial velocity $v_0 = 11.5$ m/s along flat ground, then rolls up a hill for a distance $d = 11$ m before stopping. We need to find the acceleration $a$ experienced by the ball as it rolls up the hill. 2. **Relevant formula:** Use the kinematic equation relating initial velocity, final velocity, acceleration, and displacement: $$v^2 = v_0^2 + 2ad$$ where $v$ is the final velocity, $v_0$ is the initial velocity, $a$ is acceleration, and $d$ is displacement. 3. **Apply known values:** The ball comes to rest at the top, so final velocity $v = 0$ m/s. Initial velocity $v_0 = 11.5$ m/s, displacement $d = 11$ m (uphill, so positive direction is initial velocity direction). 4. **Substitute into the formula:** $$0^2 = (11.5)^2 + 2a(11)$$ 5. **Solve for acceleration $a$:** $$0 = 132.25 + 22a$$ $$22a = -132.25$$ $$a = \frac{-132.25}{22}$$ 6. **Simplify the fraction:** $$a = \cancel{\frac{-132.25}{22}} = -6.0113636...$$ 7. **Final answer:** The acceleration experienced by the ball rolling up the hill is approximately $$a = -6.01\ \text{m/s}^2$$ This negative acceleration indicates the ball is decelerating as it moves uphill.