Speeding

Helicopter drops a package backwards…. yeah…?

A helicopter is flying horizontally at 8.3 m/s and an altitude of 20 m when a package of emergency medical supplies is ejected horizontally backward with a speed of 13 m/s relative to the helicopter. Ignoring air resistance, what is the horizontal distance between the package and the helicopter when the package hits the ground?

I disagree with the above answer, they feel that they do not need to take into account the distance the helicopter will fly during the time the package falls ….. I think that you do need to take this into account, as distances will be measured with respect to the earth ….

I believe you need to have the horiz speed measured with respect to [ wrt ] the ground ……. The package was moving 8.3 m/sec forwards with the helicopter [ which is measured wrt the ground ] ….. but then 13 m / sec backwards, wrt the helicopter [ you treat the helicopter as a stationary frame of reference ] —-> 13 – 8.3 = 4.7 m / sec backwards, wrt the ground…..

dist horiz = Vt, and since h = 0.5gt^2 ,
t = sqrt ( 2h / g) = sqrt ( 40 / 9.8 ) = 2.02 sec …. this is the time the package takes to fall to the ground ….
then d = ( 4.7 ) ( 2.02 ) = 9.5 m
from where it was released … assuming it’s horizontal velocity remains constant during it’s fall …..

If you want the distance, horizontally from the helicopter to the package as it hits the ground, then add X = vt for the helicopter ….

x = [ 8.3 ] * [ 2.02 ] = 16. 8 m , this is how far the helicopter will fly while the package falls, assuming constant velocity …

total distance = 16.8 + 9.5 = 26.3 meters , so the package should end up a total of 26.3 m behind the helicopter when it hits the ground …

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