physics
posted by Anonymous .
A passenger in a helicopter traveling upwards at 20 m/s accidentally drops a package out the window. If it takes 15 seconds to reach the ground, how high to the nearest meter was the helicopter when the package was dropped?

Neglect air friction.
Solve this equation for height, H:
Y (height above ground)
= H + 20t 4.9 t^2 = 0
using t = 15 s. 
After leaving the helicopter the package is moving upwards at the initial velocity v0.
This is decelerated motion with acceleration –g (g≈10m/s^2)
h0=v0t1(gt^2)/2
v=0=v0gt1
t1=v0/g=20/10=2 s.
After this time interval the package is at the height
h0=v0^2/2g= 20 m.
Then the package is falling down during t2=tt0=152=13 s:
H= h0+h=(gt2^2)/2=(10x169)/2=845 m.
Hence, the height of helicopter when the package was dropped is
h=Hh0=84520=825 m