The crane lowers the load vertically downward at a speed of v = 4 m / s.

The crane lowers the load vertically downward at a speed of v = 4 m / s. When the load is at a height of H = 28m, the cable breaks. What is the drop time?

V0 = 4 m / s.

g = 10 m / s2.

h = 28 m.

t -?

Since after separation, only gravity acts on the load, therefore it moves with the acceleration of gravity g.

Let’s write down the formula for acceleration: g = (V – V0) / t.

t = (V – V0) / g.

Let us express the height of the fall h by the formula: h = (V ^ 2 – V0 ^ 2) / 2 * g.

V ^ 2 – V0 ^ 2 = 2 * g * h.

V ^ 2 = V0 ^ 2 + 2 * g * h.

V = √ (V0 ^ 2 + 2 * g * h).

V = √ ((4 m / s) ^ 2 + 2 * 10 m / s2 * 28 m) = 24 m / s.

t = (24 m / s – 4 m / s) / 10 m / s2 = 2 s.

Answer: the time of the drop of the load was t = 2 s.