What Goes $Up\dots$ With the aid of calculus it is easy to show that the height $s$ of a projectile launched straight upward from an initial height $s_{\text{o }}$ with an initial velocity $v_0$ is given by $s=-\frac{1}{2}g{t}^2+v_{0}t+s_0,$ where $t$ is in seconds and $g=32\mathrm{ft}/{\mathrm{s}}^2$. If a toy rocket is shot straight upward from ground level, then $s_o=0.$ If its initial velocity is $72\mathrm{ft}/\mathrm{s}$, during what time interval will the rocket be more than $80\mathrm{ft}$ above the ground?

Key Concepts: Projectile Motion, Vertical Motion, Quadratic Equation Explanation: We are given the equation of height of a projectile shot vertically upwards as a function of time, and the initial conditions (height and velocity). Our task is to calculate the time interval when the height is greater than a given value (80ft). Step by Step Solution: Step 1. Substitute the given values into the equation s=-0.5*g*t^2 + v0*t + s0. Step 2. Rewrite the equation as -0.5*g*t^2 + v0*t + s0 - 80 > 0, since we want to find the time interval when the height is more than 80ft Step 3. Simplify this as a quadratic equation in t: -16t^2 + 72t - 80 > 0, since g = 32ft/s^2. Note that we divided all the coefficients by 2. Step 4. Solve this inequality to get the time interval. First, find the roots of the equation -16t^2 + 72t - 80 = 0. Solving the quadratic equation, we get t = 1.5s and t = 2.5s. These are the times when the height of the rocket is exactly 80ft. So the rocket will be above 80ft for the time interval (0s, 1.5s) and (2.5s, ∞), since it will hit the ground at t = 3s if not before. Step 5. Therefore, the time interval when the rocket will be more than 80ft above the ground is (0s, 1.5s) and (2.5s, ∞). We can write this as t < 1.5 or t > 2.5. Final Answer: The toy rocket will be more than 80ft above the ground during the time interval (0s, 1.5s) and (2.5s,∞), or t < 1.5 or t > 2.5 seconds.