1. **Problem Statement:** A bullet is shot upwards with an initial velocity of 200 ft/sec from a point 30 ft above the ground. Its height at time $t$ seconds is given by the quadratic function: $$h(t) = -25 t^2 + 200 t + 30$$ We need to find: (i) The maximum height the bullet reaches. (ii) The time it takes to reach this maximum height. 2. **Formula and Explanation:** The height function $h(t)$ is a quadratic function of the form: $$h(t) = at^2 + bt + c$$ where $a = -25$, $b = 200$, and $c = 30$. Since $a < 0$, the parabola opens downward, so the vertex represents the maximum point. The time $t$ at which the maximum height occurs is given by the vertex formula: $$t = -\frac{b}{2a}$$ 3. **Calculate the time to reach maximum height:** $$t = -\frac{200}{2 \times (-25)} = -\frac{200}{-50} = 4$$ seconds. 4. **Calculate the maximum height:** Substitute $t=4$ into $h(t)$: $$h(4) = -25(4)^2 + 200(4) + 30 = -25(16) + 800 + 30 = -400 + 800 + 30 = 430$$ feet. 5. **Summary:** - The bullet reaches its highest point at $t=4$ seconds. - The maximum height is $430$ feet. This means the bullet will rise to 430 feet above the ground after 4 seconds before starting to fall back down.