Question

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?

Answer 1

check the picture below

so
`\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)`

so it reaches a maximum height of
`\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}} \quad feet` and that happens at
`\bf -\cfrac{{{ b}}}{2{{ a}}} \quad seconds`

Answer 2

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