1. **State the problem:** We have a toy rocket launched from a platform 48 feet high, with height modeled by the equation $$h = -16t^2 + 32t + 48$$ where $h$ is height in feet and $t$ is time in seconds. 2. **Part A: Find the height after 1 second.** Substitute $t=1$ into the equation: $$h = -16(1)^2 + 32(1) + 48$$ $$h = -16 + 32 + 48$$ $$h = 64$$ So, the height after 1 second is 64 feet. 3. **Part B: Find the time when the rocket reaches the ground.** The rocket reaches the ground when $h=0$: $$0 = -16t^2 + 32t + 48$$ Divide both sides by -16 to simplify: $$0 = \cancel{-16}t^2 + \cancel{-16}\left(-2\right)t + \cancel{-16}\left(-3\right)$$ $$0 = t^2 - 2t - 3$$ Now solve the quadratic equation: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-2$, $c=-3$. Calculate the discriminant: $$\sqrt{(-2)^2 - 4(1)(-3)} = \sqrt{4 + 12} = \sqrt{16} = 4$$ Find the roots: $$t = \frac{2 \pm 4}{2}$$ Two solutions: $$t = \frac{2 + 4}{2} = 3$$ $$t = \frac{2 - 4}{2} = -1$$ Since time cannot be negative, the rocket hits the ground at $t=3$ seconds. **Final answers:** - Part A: 64 feet - Part B: 3 seconds