At a particular instant, a stationary observer on the ground sees a package falling with speed v1 at an angle to the vertical. To a pilot flying horizontally at constant speed relative to the ground, the package appears to be falling vertically with a speed v2 at that instant. What is the speed of the pilot relative to the ground?
Ok, so the object falling relative to the observer has a different velocity then it does relative to the pilot. The y-component of the objects velocity will be identical for both reference frames, but the horizontal velocity is going to differ.
Ground speed is equal to the vector sum of Airspeed and Windspeed. If the pilot observes the package to be falling vertically, he must be approaching it from behind. Assuming the package was dropped vertically, the x-component of its velocity should reflect the windspeed?
So, the ground speed of the plane would be the sums of its constant velocity and the x-component of the objects velocity?
I am asking for some guidance with this questions, but not the answer, if someone could give me some hints.
Best Answer
You already have all the points needed. I'll quote you:
and then you say
Now what you should do is to think of a famous theorem that connects the sum of the squares of the components of a vector to its magnitude. The answer will pop out then..!