21) An object in launched upward at 64 feet per second (ft/s) from a platform 80 feet high. The equation for a) When does the object land on the ground? Tip: These questions are always going to ask you for the $x$-intercepts and vertex. Ground $\rightarrow x$-int b) What is the max height of the object? Max or Min $\rightarrow$ vertex c) When is the object 100 feet off the ground?

Step 1 ：Given the initial velocity of 64 ft/s and initial height of 80 feet, the equation for the height of the object as a function of time is: \(h(t) = -16t^2 + 64t + 80\)

Step 2 ：a) To find when the object lands on the ground, we need to find the time t when \(h(t) = 0\): \(-16t^2 + 64t + 80 = 0\). The object lands on the ground at \(t = 5\) seconds.

Step 3 ：b) To find the max height of the object, we need to find the vertex of the parabola represented by the equation \(h(t)\). The vertex occurs at \(t = \frac{-b}{2a} = \frac{-64}{2(-16)} = 2\) seconds. The max height is \(h(2) = -16(2)^2 + 64(2) + 80 = 144\) feet.

Step 4 ：c) To find when the object is 100 feet off the ground, we need to find the time t when \(h(t) = 100\): \(-16t^2 + 64t + 80 = 100\). The object is 100 feet off the ground at \(t = 2 - \frac{\sqrt{11}}{2}\) and \(t = 2 + \frac{\sqrt{11}}{2}\) seconds.