A rescue plane flying horizontally at 72.6 m/s spots a survivor in the ocean 182 m directly below and releases an emergency kit with a parachute. Because of the shape of the parachute, it experiences insignificant horizontal air resistance . If the kit descends with a constant vertical acceleration of 5.82 m/s^2, how far away from the survivor will it hit the waves? Show work please....

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The kit should travel 182m vertical distance with an acceleration of 5.82m/s^2. Here we have to assume that the kit was released at rest.

Using equations of motion;

`darrs = ut+1/2at^2`

`182 = 0+1/2xx5.82xxt^2`

`t = 7.9s`

So with in 7.9 seconds the kit reach to sea. At the moment...

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The kit should travel 182m vertical distance with an acceleration of 5.82m/s^2. Here we have to assume that the kit was released at rest.

Using equations of motion;

`darrs = ut+1/2at^2`

`182 = 0+1/2xx5.82xxt^2`

`t = 7.9s`

So with in 7.9 seconds the kit reach to sea. At the moment of release it has a horizontal velocity component due to the velocity of plane which is 72.6m/s

`rarr s = ut+1/at^2`

`s = 72.6xx7.9 = 574.15m`

So the kit will hit the sea 574.15m away from the survivor.

Note

Any way this answer is not pratical at all. It is hard to believe that survivor can travel 500m or more to get a emergency kit.

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