1. Sep 21, 2008

A helicopter resting on ground starts to accelerate uniformly upwards. After 30s a flare was fired horizontally from the helicopter. 10s later, the flare reached the ground. (g = 9.81m/s^2).
Calculate:

a)Acceleration of helicopter
b)Height from which flare was fired
c)Height of helicopter the instant flare reaches ground (assuming it continued its constant acceleration after flare was fired)

*****I have already been told that i must show my work before getting help but that is why i am here :(. I do not want the answer, i just want someone to point me in the right direction.

Thank you

2. Sep 21, 2008

### Almanzo

After 10 seconds, the flare will have a vertical speed (vertical component of velocity) of 10 s * 9.81 m/s2 = 98. 1, m/s. (Its vertical speed was initially zero, as it was fired in a horizontal direction. Unless they mean horizontal relative to the ascending helicopter. I assume they don't mean that.)

The flare's mean vertical speed during its fall will be half of this. Multiplying this mean speed with 10 seconds will give the (vertical) distance covered, i.e. the height of the helicopter when the flare was fired.

That height was attained after thirty seconds. This allows us to calculate the mean speed of ascension during those thirty seconds. The actual speed after thirty seconds will be double the mean speed (because the helicopter was uniformly accelerating upwards). Dividing this new speed by thirty seconds, we find the acceleration. Now we can multiply this acceleration by forty seconds (30 + 10) and find the speed of the helicopter when the flare hits the ground. Dividing by two gives the mean speed of the helicopter during those forty seconds. Multiplying this mean speed by forty seconds should give us the height of the helicopter.

3. Mar 7, 2011