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A package is dropped from a helicopter falling at 36 m/s.

If it takes 11.0s before the package strikes the ground, how high above the ground was the package when it was released? Ignore air resistance

User Bora
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2 Answers

3 votes

396 meters. because if the package is falling at 36m/s then, we just have to multiply 36×11 and we get our height which is 396 meters bb:)

User Spydon
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3 votes

Answer:

The height above the ground is 988.9 meters.

Step-by-step explanation:

It is given that,

Initial speed of the package, u = 36 m/s

Time taken by the package to reach the ground, t = 11 s

Let h is the height above the ground when it was released. It can be calculated using the second equation of motion as :


h=ut+(1)/(2)at^2

a = g


h=ut+(1)/(2)gt^2


h=36* 11+(1)/(2)* 9.8* 11^2

h = 988.9 meters

So, the height above the ground is 988.9 meters. Hence, this is the required solution.

User Scott Nicol
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