1. **State the problem:** We have an object shot vertically from a 100 m high cliff with initial speed 274.4 m/s. Its height at time $t$ seconds is given by $$y = -4.9t^2 + 274.4t + 100.$$ We need to find: a. The time and value of the maximum height. b. The total flight time before it hits the ocean (height 0). c. The height at $t=8$ seconds. d. The two times when the height is 2525.5 m. 2. **Formula and rules:** This is a quadratic function in standard form $y = at^2 + bt + c$ with $a = -4.9$, $b = 274.4$, and $c = 100$. - The vertex (maximum point since $a<0$) occurs at $$t = -\frac{b}{2a}.$$ - The maximum height is $$y_{max} = y(t)$$ at this $t$. - The object hits the ocean when $y=0$, solve $$-4.9t^2 + 274.4t + 100 = 0$$ for $t$. - To find height at $t=8$, substitute $t=8$ into the equation. - To find times when $y=2525.5$, solve $$-4.9t^2 + 274.4t + 100 = 2525.5.$$ 3. **Find maximum height and time:** $$t = -\frac{274.4}{2 \times (-4.9)} = \frac{274.4}{9.8} = 28$$ seconds. Substitute $t=28$: $$y = -4.9(28)^2 + 274.4(28) + 100 = -4.9(784) + 7683.2 + 100 = -3831.6 + 7683.2 + 100 = 3951.6$$ meters. 4. **Find total flight time (when $y=0$):** Solve $$-4.9t^2 + 274.4t + 100 = 0.$$ Use quadratic formula: $$t = \frac{-274.4 \pm \sqrt{(274.4)^2 - 4(-4.9)(100)}}{2(-4.9)}.$$ Calculate discriminant: $$274.4^2 = 75299.36,$$ $$4 \times -4.9 \times 100 = -1960,$$ So, $$\sqrt{75299.36 + 1960} = \sqrt{77259.36} \approx 278.04.$$ Then, $$t = \frac{-274.4 \pm 278.04}{-9.8}.$$ Two roots: $$t_1 = \frac{-274.4 + 278.04}{-9.8} = \frac{3.64}{-9.8} = -0.37 ext{ (discard negative time)},$$ $$t_2 = \frac{-274.4 - 278.04}{-9.8} = \frac{-552.44}{-9.8} = 56.36 ext{ seconds}.$$ So, the object hits the ocean after approximately 56.36 seconds. 5. **Height at $t=8$ seconds:** $$y = -4.9(8)^2 + 274.4(8) + 100 = -4.9(64) + 2195.2 + 100 = -313.6 + 2195.2 + 100 = 1981.6$$ meters. 6. **Times when height is 2525.5 m:** Solve $$-4.9t^2 + 274.4t + 100 = 2525.5,$$ or $$-4.9t^2 + 274.4t + 100 - 2525.5 = 0,$$ $$-4.9t^2 + 274.4t - 2425.5 = 0.$$ Use quadratic formula: $$t = \frac{-274.4 \pm \sqrt{274.4^2 - 4(-4.9)(-2425.5)}}{2(-4.9)}.$$ Calculate discriminant: $$274.4^2 = 75299.36,$$ $$4 \times -4.9 \times -2425.5 = 4 \times 4.9 \times 2425.5 = 47519.8,$$ So, $$\sqrt{75299.36 - 47519.8} = \sqrt{27779.56} \approx 166.67.$$ Then, $$t = \frac{-274.4 \pm 166.67}{-9.8}.$$ Two roots: $$t_1 = \frac{-274.4 + 166.67}{-9.8} = \frac{-107.73}{-9.8} = 11.0 ext{ seconds},$$ $$t_2 = \frac{-274.4 - 166.67}{-9.8} = \frac{-441.07}{-9.8} = 45.0 ext{ seconds}.$$ **Final answers:** - a. Maximum height is 3951.6 m reached at 28 s. - b. Flight time before hitting ocean is 56.36 s. - c. Height at 8 s is 1981.6 m. - d. Height 2525.5 m occurs at 11.0 s and 45.0 s after release.