An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers. The plane is traveling horizontally at 40.0 m/s at a height of 100 m above the ground.

Question

a) How far horizontally from the dropping point does the package strike the ground?
b)What are the x and y velocities before it hits the ground?

Answer 1

s = ut + 1/2 at^2.............since the initial velocity down u = 0

100 = 0 + 1/2(9.81)t^2

t = 4.51 sec

again s = ut + 1/2 at^2...........this time horizontally

s = distance = (40)(4.51) + 1/2(0)(4.51)^2................Notice the acceleration = 0

s = 180.61 m away in the direction of flight

this is the same as saying distance = speed x time.

II. Kinetic Energy = 1/2 mv^2

Potential Energy = mgh

(1/2)m(40)^2 + m(9.81)(100) = 1/2 m v^2............................

divide by m

(1/2) (1600) + 981 = 1/2(v^2)

v = 59.6 m/s

This ia according to the law of conservation of Energy.

Another way to do it is Horizontal and vertical velocity:

Horizontal v = 40 m/s...............since a = 0

Vertical v = u + at = 0 + (9.81)(4.51) = 44.23 m/s

Using pythagoras the resulting v = Squaroot ( 44.23^2 + 40^2 ) = 59.6 m/s

Same answer......and it should be, either way would do..