An Alaskan rescue plane traveling 52 m/s drops a package of emergency rations from a height of 115 m to a stranded paty of explorers. The acceleration of gravity is 9.81 m/s2. Where does the package strike the ground relative to the point directly below where it was released? What is the horizontal component of the velocity just before it hits? What is the vertical component of the velocity just before it hits?
Ignoring air friction, the horizontal velocity DOES NOT CHANGE. There is NO horizontal force, therefore NO horizontal acceleration. That is what you are supposed to get :)
( If you drop a bomb out of an airplane, turn or it will hit right under you)
Vertical problem
a = -9.81
Vi = initial speed down = 0
Hi = initial height = 115
v = 0 - 9.81 t
h = 115 - 0 t - 4.9 t^2
h = 0 at ground
0 = 115 - 4.9 t^2
t^2 = 23.5
t = 4.84 seconds to ground
v = -9.81 * 4.84 = -47.5 meters/ second vertical speed down at ground
HORIZONTAL PROBLEM (right under the plane :)
u = 52 m/s
t = 4.84
x = u * t = 252 meters