Apackage is dropped from a helicopter that is moving upward at 15 m/s.
if it takes 8.0s before the package strikes the ground, how high above the ground was the package when it was released?

193.6 m.

Explanation:

Consider upward direction as negative and downward direction as positive. Initial velocity in vertical velocity of the package = -15 m/s

Time taken by the package to reach the ground = 8.0 s

Acceleration due to gravity will act in downward direction.

Use the second equation of motion:

s = u t + 0.5 at²

where, s is the displacement, u is the initial velocity, t is the time and a is the acceleration.

Substitute the values:

s = (-15)(8)+ 0.5 (9.8)(8.0)² = 193.6 m

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.

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