1. **Problem Statement:** We are given the height of a ball during a slam dunk modeled by the quadratic function: $$h(t) = -5t^2 + 10t + 4$$ where $h(t)$ is the height in meters and $t$ is the time in seconds. We need to: - Find the maximum height (vertex) of the ball. - Determine the total time taken for the ball to return to the basket when $h(t) = 3$ meters. - Modify the equation for higher and lower jump velocities and analyze the changes. 2. **Finding the Maximum Height (Vertex):** The height function is a quadratic of the form: $$h(t) = at^2 + bt + c$$ where $a = -5$, $b = 10$, and $c = 4$. The vertex time $t_{vertex}$ is given by: $$t_{vertex} = -\frac{b}{2a} = -\frac{10}{2 \times (-5)} = 1$$ Substitute $t=1$ into $h(t)$ to find the maximum height: $$h(1) = -5(1)^2 + 10(1) + 4 = -5 + 10 + 4 = 9$$ **Maximum height is 9 meters at 1 second.** 3. **Finding Total Time When $h(t) = 3$ meters:** Set height equal to 3: $$-5t^2 + 10t + 4 = 3$$ Simplify: $$-5t^2 + 10t + 1 = 0$$ Multiply both sides by -1 for easier calculation: $$5t^2 - 10t - 1 = 0$$ Use quadratic formula: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=5$, $b=-10$, $c=-1$. Calculate discriminant: $$\Delta = (-10)^2 - 4 \times 5 \times (-1) = 100 + 20 = 120$$ Calculate roots: $$t = \frac{10 \pm \sqrt{120}}{10} = \frac{10 \pm 2\sqrt{30}}{10} = 1 \pm \frac{\sqrt{30}}{5}$$ Approximate values: $$t_1 = 1 - 1.095 = -0.095 \text{ (discard negative time)}$$ $$t_2 = 1 + 1.095 = 2.095$$ **Total time for ball to return to height 3 meters is approximately 2.095 seconds.** 4. **Modifying the Equation for Different Jump Velocities:** The coefficient of $t$ (which is 10) represents the initial velocity component. - For a higher jump, increase this coefficient, e.g., 14. - For a lower jump, decrease this coefficient, e.g., 6. New equations: - Higher jump: $$h_1(t) = -5t^2 + 14t + 4$$ - Lower jump: $$h_2(t) = -5t^2 + 6t + 4$$ 5. **Effect of Changes on Motion:** - Increasing the coefficient increases the initial velocity, resulting in a higher maximum height and longer time in the air. - Decreasing the coefficient lowers the maximum height and reduces the time the ball stays in the air. 6. **Summary:** - Original max height: 9 m at 1 s. - Time to return to 3 m: approx 2.095 s. - Higher jump max height and time increase. - Lower jump max height and time decrease.