Acceleration of the helicopter as it starts to rise

[anlenemilk]

Homework Statement

At take off, the rotor blades of a helicopter propel 2400 kg of air vertically downwards each second. The air, intially at rest, is given a speed of 11ms^-1
QN: the mass of the helicopter is 2500 kg. Find the acceleration of the helicopter as it starts to rise

Homework Equations

<F>=Change in momentum /change in time
Fnet=ma

The Attempt at a Solution

<F>=Change in momentum /change in time
= (2400*11-0)/1= 26400N
26400-2500(9.81)=(2500)a
a=0.75ms^-2

The answer is 9.58ms^-2
I have no idea how to get this answer! HELP ME! THANK YOU VERY MUCH

 

[Doc Al]

I'd say that your answer is correct and the given answer is wrong.

What textbook is this problem from?

 

[Doc Al]

26400-2500(9.81)=(2500)a

Note that you'll get the given answer if you changed one of the masses to 250 (which is clearly wrong):

26400-250(9.81)=(2500)a​

That could be how the wrong answer was attained.