# No 7: Two bodies of mass `m_1` and `m_2` are dropped from rest from the heights `h_1` and `h_2` . What will be the ratio of the times they take to reach the ground? The answer choices are shown on the image below.

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This image has been Flagged as inappropriate Click to unflag The answer is b: `sqrt(h_1) : sqrt(h_2)` .

If a body is dropped from rest from height h, the time it takes to reach the ground is determined by the equation

`h = g*t^2/2` , where g is the gravitational acceleration, g = 9.8 m/s^2. Note that the mass of the body is not included in this equation. Even if the bodies have different mass, when they are dropped from the same height, they will reach the ground at the same time (assuming the air resistance is negligible.)

Applying this equation to the two bodies results in

`h_1 = g*t_1^2/2` , from where `t_1 = sqrt((2h_1)/g)` and

`h_2 = g*t_2^2/2` , from where `t_2 = sqrt((2h_2)/g)` .

Notice that the times are proportional to the square roots of the heights.

The ratio of the times will be

`t_1 : t_2 = sqrt((2h_1)/g) : sqrt((2h_2)/g) = sqrt(h_1) : sqrt(h_2)` , so the answer is choice b.

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