Acceleration of the helicopter as it starts to rise!

  • Thread starter anlenemilk
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  • #1
anlenemilk
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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
 

Answers and Replies

  • #2
Doc Al
Mentor
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I'd say that your answer is correct and the given answer is wrong.

What textbook is this problem from?
 
  • #3
Doc Al
Mentor
45,410
1,844
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.
 

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