Ajet with mass m = 9.0 kg jet accelerates down the runway for takeoff at 1.6 m/s2.

3) once off the ground, the plane climbs upward for 20 seconds. during this time, the vertical speed increases from zero to 24.0 m/s, while the horizontal speed increases from 80.0 m/s to 98.0 m/s.

5) after reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 13.0 seconds.

what is the net vertical force on the airplane as it levels off?

This question requires the use of the equation of motion:
v = u + at [v is final velocity (0), u is initial velocity (24), a is acceleration, t is time (13)]
to calculate the acceleration. This can then be multiplied by the mass of the plane to obtain the net force via:
F = ma (F is force, m is mass, a is acceleration)
First, we calculate the acceleration:
0 = 24 + 13(a)
a = -24/13 m/s^2
The force is then:
F = 90000 * (-24/13)
F = -1.66*10^5 Newtons
The negative sign indicates that the force and acceleration are in the opposite direction as the velocity (since we took velocity to be positive)