A model airplane is shot up from a platform 1 foot above the ground with an initial upward velocity of 56 feet per second. The height of the airplane above ground after t seconds is given by the equation h=-16t^2+56t+1 where h is the height of the airplane in feet and t is the time in seconds after it is launched. Approximately how long does it take the airplane to reach its maximum height?
h=-16t^2+56t+1, the airplane is shot up from a platform 1 foot above the ground, so 1= - 16t^2+56t+1, we solve this equation (- 16t+56)t=0, (- 16t+56)=0 or t=o, - 16t+56=0, implies t =3.5s so the answer is C)3.5 seconds
so [tex]\bf \textit{vertex of a parabola}\\ \quad \\
\begin{array}{lccclll}
h=&-16t^2&+56t&+1\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)[/tex]
so it reaches a maximum height of [tex]\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}} \quad feet[/tex] and that happens at [tex]\bf -\cfrac{{{ b}}}{2{{ a}}} \quad seconds[/tex]
