Question

An object is dropped from 34 feet below the trip of the pinnacle atop a 710 fr tall building. The height h if the object after t seconds is given the equation h=-16t to the power of 2 +676 find how many seconds pass baffle the object reaches the ground

Answer 1

Setting h=0 and solving for t we get:

`\begin{gathered} 0=-16t^2+676, \\ 16t^2=676, \\ t^2=(676)/(16), \\ t^2=42.25 \\ t=\pm\sqrt[]{42.25}. \end{gathered}`

Since t<0 does not have real meaning, we get that:

`t=6.5.`

Answer: 6.5 seconds.