A 4.0 kg food pack was dropped from a height of 200 m by a helicopter during flood relief operations. At what velocity will the food pack be falling halfway to the ground?

Explanation:

Given:

Distance, s = 100 m

Initial velocity, u = 0 m/s

Acceleration, a = 9.8 m/s^2

Using equations of motion,

V^2 = u^2 + 2as

V = sqrt(2 × 9.8 × 100)

= 44.47 m/s

Velocity at halfway down to the ground is 44.45 m/s and as tge food package keeps going down(dustance to the ground reduces), the velocity increases.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

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