A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released? (Disregard...

A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released? (Disregard air resistance.)

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Note: The answer is given it's should be 1000 m ??

Asked on by esila1

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llltkl | College Teacher | (Level 3) Valedictorian

Posted on

The motion of the package can be broken down into two parts:

1) The package’s upward motion due to its original upward velocity, and then

2) Its free fall due to gravity from the highest point of its flight.

For the upward journey, the equations of motion are:

`v_y=u-g*t`

`S_y=h+ut-1/2g*t^2` (where h is the initial height from where the package was dropped)

At the highest point of its flight, `v_y=0`

`rArr t=u/g=15/9.81=1.53 s`

So, `S_y=h+( 15*1.53-1/2*9.81*1.53^2)`

=`(h+11.5)` m

For the free fall, the equations of motion are:

`v_y=g*t`

`S_y=1/2g*t'^2` (where t’ is the time of downward travel =16-1.53=14.47 s)

`=1/2*9.81*14.47^2`

`=1027 ` m

Therefore, `h=S_y-11.5=1027-11.5=1015.5` m

`~~1000` m

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