1. **State the problem:** We have the height of a javelin as a function of time $t$ given by $$h = -2t^2 + 9t + 11.$$ We need to find: (a) The time when the javelin hits the ground (i.e., when $h=0$). (b) The time when the javelin reaches its maximum height. (c) The maximum height reached by the javelin. 2. **Formula and rules:** - The javelin hits the ground when $h=0$. - The height function is a quadratic with $a=-2$, $b=9$, and $c=11$. - The time at maximum height is the vertex of the parabola, given by $$t = -\frac{b}{2a}.$$ - The maximum height is the value of $h$ at this $t$. 3. **(a) Find when the javelin hits the ground:** Set $h=0$: $$-2t^2 + 9t + 11 = 0.$$ Divide both sides by $-1$ to simplify: $$2t^2 - 9t - 11 = 0.$$ Use the quadratic formula: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-9$, $c=-11$. Calculate the discriminant: $$\Delta = (-9)^2 - 4 \times 2 \times (-11) = 81 + 88 = 169.$$ Calculate the roots: $$t = \frac{-(-9) \pm \sqrt{169}}{2 \times 2} = \frac{9 \pm 13}{4}.$$ Two solutions: $$t_1 = \frac{9 + 13}{4} = \frac{22}{4} = 5.5,$$ $$t_2 = \frac{9 - 13}{4} = \frac{-4}{4} = -1.$$ Since time cannot be negative, the javelin hits the ground at $$t = 5.5$$ seconds. 4. **(b) Find the time when the javelin reaches maximum height:** Use vertex formula: $$t = -\frac{b}{2a} = -\frac{9}{2 \times (-2)} = -\frac{9}{-4} = 2.25.$$ So, the javelin reaches maximum height at $$t = 2.25$$ seconds. 5. **(c) Find the maximum height:** Substitute $t=2.25$ into the height equation: $$h = -2(2.25)^2 + 9(2.25) + 11.$$ Calculate: $$-2(5.0625) + 20.25 + 11 = -10.125 + 20.25 + 11 = 21.125.$$ So, the maximum height is $$21.125$$ meters. **Final answers:** (a) $5.5$ seconds (b) $2.25$ seconds (c) $21.125$ meters