Posted by **Anonymous** on Thursday, February 23, 2012 at 8:18am.

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?

- physics -
**drwls**, Thursday, February 23, 2012 at 8:58am
Neglect air friction.

Solve this equation for height, H:

Y (height above ground)

= H + 20t -4.9 t^2 = 0

using t = 15 s.

- physics -
**Elena**, Thursday, February 23, 2012 at 9:29am
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=v0-gt1

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=t-t0=15-2=13 s:

H= h0+h=(gt2^2)/2=(10x169)/2=845 m.

Hence, the height of helicopter when the package was dropped is

h=H-h0=845-20=825 m

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