# Dog Sledding

A cargo plane is dropping supplies to a group of people dogsledding acoss Antarctica. The plance drops the package when it is directly above the group of people at an altitude of 1950m. The dogsled travels at 5.68m/s and reaches the supplies in 6.70mins. What was the speed of the cargo plane?

well I`m assuming that the plane and dogsled are traveling in opposite directions. So lets make the dogsled travel in the (+)ve direction and the plane traveling in the (-)ve direction.

Xo = o
X = ??
to = 0s
t = 402s
Vdogsled = 5.68m/s
Vplane = ???

and then I used this equation: X = Xo + Vot to determine X
X = 0 + 5.68m/s(402s)
X = 2283.36m

then using tan^-1 (theta) I found the angle which is 40.50degrees

but now I'm stuck not quite sure what to do next - not quite sure if I'm on the right tracl :uhh:

I think you are on the right track. But why would you assume the plane and sleddge are travelling in different directions?

From the point you are now, what if you calculated how long it takes for the package to dropp to the ground? I think you can assume free fall.

alright - I'll try that

HallsofIvy