Answer
The takeoff speed is 73 m/s.
Work Step by Step
$ma = F_T ~sin(\theta)$
$mg = F_T ~cos(\theta)$
We can divide the first equation by the second equation:
$\frac{a}{g} = tan(\theta)$
$a = g ~tan(\theta) = (9.80 ~m/s^2) ~tan(25^{\circ})$
$a = 4.57 ~m/s^2$
Now, we use the acceleration to find the takeoff speed:
$v = v_0 + at = 0 + at = (4.57 ~m/s^2)(16 ~s)$
$v = 73 ~m/s$
The takeoff speed is 73 m/s.
