Two teams of movers are lowering a piano from the window of a 10 floor apartment building. The rope breaks when the piano is 30 meters above the ground. The movers on the ground, alerted by the shouts of the movers above, first notice the piano when it is 14 meters above the ground. How long do they have to get out of the way before the piano hits the ground

The only force acting on the piano will be the gravitational one, thus the acceleration of the piano is the gravitational acceleration: a(t) = -9.8m/s^2 (Where the negative sign is because it is falling down).
To get the velocity we integrate over time, because the piano has no initial velocity, the constant of integration is zero: v(t) = (-9.8m/s^2)*t
To get the position equation we integrate again, here the initial position is 30 meters above the ground, so that will be our constant of integration: p(t) = (1/2)*(-9.8m/s^2)*t^2 + 30m

We want to find the value of t such that the position is equal to zero, so we need to solve:

0 = (1/2)*(-9.8m/s^2)*t^2 + 30m

30m = (1/2)*(9.8m/s^2)*t^2

2*30m/(9.8m/s^2) = t^2

6.1 s^2 = t^2

√(6.1 s^2) = 2.5s = t

This means that the piano falls to the ground in 2.5 seconds.

But the workes notice it when the piano is 14 meters above the ground, it happens when:

p(t) = 14m = (1/2)*(-9.8m/s^2)*t^2 + 30m

Solving that we get:

30m - 14m = (1/2)*(9.8m/s^2)*t^2

2*16m/(9.8m/s^2) = t^2

3.27s^2 = t^2

√(3.27s^2) = t = 1.8s

So the piano falls in 2.5 seconds, and the works notice it 1.8 seconds after it starts falling, meaning that they have:

2.5 - 1.8 = 0.7 seconds to react.

If you want to learn more about motion equations, you can read:

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