1. **State the problem:** We have a cannon ball launched with an initial velocity of 56 feet per second. Its height after $t$ seconds is given by the equation: $$h = -16t^2 + 56t + 7$$ We need to find: A) The initial height of the cannon ball. B) The time it takes for the cannon ball to hit the ground (when height $h=0$). 2. **Formula and rules:** - The initial height is the height at time $t=0$. - To find when the cannon ball hits the ground, solve $h=0$ for $t$. - The equation is a quadratic in standard form $at^2 + bt + c = 0$ with $a=-16$, $b=56$, and $c=7$. - Use the quadratic formula: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Find the initial height:** Substitute $t=0$ into the height equation: $$h = -16(0)^2 + 56(0) + 7 = 7$$ So, the initial height is 7 feet. 4. **Find when the cannon ball hits the ground:** Set $h=0$: $$0 = -16t^2 + 56t + 7$$ Apply the quadratic formula: $$t = \frac{-56 \pm \sqrt{56^2 - 4(-16)(7)}}{2(-16)}$$ Calculate the discriminant: $$56^2 = 3136$$ $$-4(-16)(7) = 448$$ $$\sqrt{3136 + 448} = \sqrt{3584}$$ Simplify the square root: $$\sqrt{3584} \approx 59.866$$ Now substitute back: $$t = \frac{-56 \pm 59.866}{-32}$$ Calculate both solutions: For $+$: $$t = \frac{-56 + 59.866}{-32} = \frac{3.866}{-32} = -0.121$$ (discard negative time) For $-$: $$t = \frac{-56 - 59.866}{-32} = \frac{-115.866}{-32} = 3.621$$ 5. **Final answers:** A) Initial height = 7 feet B) Time to hit the ground = 3.621 seconds (rounded to 3 decimal places)