1. **Problem statement:** A ball is thrown vertically with height given by the function $$y=16.6t-4.9t^2$$ meters after $$t$$ seconds. We need to find: a) Velocity and acceleration at time $$t$$. b) Time to reach the highest point. c) Maximum height. d) Time(s) when the ball is at half its maximum height. 2. **Formulas and rules:** - Velocity $$v(t)$$ is the first derivative of height $$y(t)$$ with respect to time $$t$$: $$v(t) = \frac{dy}{dt}$$. - Acceleration $$a(t)$$ is the derivative of velocity or the second derivative of height: $$a(t) = \frac{d^2y}{dt^2}$$. - The highest point occurs when velocity $$v(t) = 0$$. - To find time(s) at half maximum height, solve $$y(t) = \frac{1}{2} y_{max}$$. 3. **Calculate velocity:** $$v(t) = \frac{d}{dt}(16.6t - 4.9t^2) = 16.6 - 9.8t$$ 4. **Calculate acceleration:** $$a(t) = \frac{d}{dt}(16.6 - 9.8t) = -9.8$$ (constant acceleration due to gravity, downward). 5. **Find time to reach highest point:** Set velocity to zero: $$0 = 16.6 - 9.8t$$ Solve for $$t$$: $$t = \frac{16.6}{9.8} \approx 1.6939 \text{ seconds}$$ 6. **Find maximum height:** Substitute $$t = 1.6939$$ into $$y(t)$$: $$y_{max} = 16.6(1.6939) - 4.9(1.6939)^2$$ Calculate: $$y_{max} \approx 28.11 - 14.05 = 14.06 \text{ meters}$$ 7. **Find time(s) when height is half maximum:** Set $$y(t) = \frac{1}{2} y_{max} = 7.03$$: $$16.6t - 4.9t^2 = 7.03$$ Rearranged: $$-4.9t^2 + 16.6t - 7.03 = 0$$ Multiply both sides by -1 for clarity: $$4.9t^2 - 16.6t + 7.03 = 0$$ Use quadratic formula: $$t = \frac{16.6 \pm \sqrt{16.6^2 - 4 \times 4.9 \times 7.03}}{2 \times 4.9}$$ Calculate discriminant: $$16.6^2 = 275.56$$ $$4 \times 4.9 \times 7.03 = 137.788$$ $$\sqrt{275.56 - 137.788} = \sqrt{137.772} \approx 11.74$$ Calculate roots: $$t_1 = \frac{16.6 - 11.74}{9.8} = \frac{4.86}{9.8} \approx 0.496$$ $$t_2 = \frac{16.6 + 11.74}{9.8} = \frac{28.34}{9.8} \approx 2.89$$ **Answer:** a) Velocity: $$v(t) = 16.6 - 9.8t$$ m/s, Acceleration: $$a(t) = -9.8$$ m/s² b) Time to highest point: $$t \approx 1.69$$ seconds c) Maximum height: $$y_{max} \approx 14.06$$ meters d) Times at half max height: $$t \approx 0.50$$ seconds and $$t \approx 2.89$$ seconds