Hi guys, got an interesting question in my textbook ive been struggling with:

with a Westerly wind @ 8.5km/h, helicopter takes off vertically from a level field at full power. It has a mass of 700g and 10N of thrust. Once it reaches a height of 5m, stays at same and flies north. 10s later, the helicopter stops. How far away and in what direction is that point from the place where the helicopter took off?

I figured that A=14.28, and the distance it would travel in 10s is 599.54m, but im struggling to figure out how you integrate the 10km/h westerly vector into the equation.

Question

Hi guys, got an interesting question in my textbook ive been struggling with:

with a Westerly wind @ 8.5km/h, helicopter takes off vertically from a level field at full power. It has a mass of 700g and 10N of thrust. Once it reaches a height of 5m, stays at same and flies north. 10s later, the helicopter stops. How far away and in what direction is that point from the place where the helicopter took off?

I figured that A=14.28, and the distance it would travel in 10s is 599.54m, but im struggling to figure out how you integrate the 10km/h westerly vector into the equation.

doc.brown · Answer

You've got two vectors there, how far up, and how far away over the ground.
How far away over the ground is made up of two vectors, how much North, and how much West.  The wind is always blowing and affecting the position of the helicopter.