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

Respuesta :

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:)

Answer:

The height above the ground is 988.9 meters.

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 :

[tex]h=ut+\dfrac{1}{2}at^2[/tex]

a = g

[tex]h=ut+\dfrac{1}{2}gt^2[/tex]

[tex]h=36\times 11+\dfrac{1}{2}\times 9.8\times 11^2[/tex]

h = 988.9 meters

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