Question

A ball is thrown straight up from the top of a building that is 400 ft high with an initial velocity of 64 ft/s. The height of the object can be modeled by the equation s ( t ) = -16 t2 + 64 t + 400.

In two or more complete sentences explain how to determine the time(s) the ball is higher than the building in interval notation.

Answer 1

Check the picture below.

the ball is higher than the building when y = 0, or namely when s(t) = 0, any other time is above it, and hmmm when it's that?

`\stackrel{s(t)}{0}~~ = ~~-16t^2+64t+400\implies 0=-16(t^2-4t-25) \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{1}t^2\stackrel{\stackrel{b}{\downarrow }}{-4}t\stackrel{\stackrel{c}{\downarrow }}{-25} \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t= \cfrac{ - (-4) \pm \sqrt { (-4)^2 -4(1)(-25)}}{2(1)}\implies t=\cfrac{4\pm√(116)}{2}`

`t=\cfrac{4\pm 2√(29)}{2}\implies t= \begin{cases} 2+√(29)~~\textit{\Large \checkmark}\\ 2-√(29) \end{cases}~\hfill \boxed{(0~~,~~2√(29))}`

notice, we can't use the 2-√29 because is a negative value and "t" is never negati ve, also notice the interval notation, we have parentheses on both sides.