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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