On Mars a rock falls an unknown vertical distance from a resting position and lands in a crater. If it takes the rock 2.5 seconds to fall, how high is the cliff the rock fell from? Mars' surface gravity is 3.8 m/s squared.

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Denote the gravity acceleration near the surface of Mars as `g_M.` The only force acting on a rock is the gravity force which gives a rock a constant acceleration of `g_M` by Newton's Second law. We can ignore air resistance because the atmosphere of Mars is not very dense and the speeds are moderate.

The initial speed is zero ("falls from a resting position"). Therefore the height of a falling rock may be determined by the equation

`H(t)=H_0-g_M*t^2/2,`

where `H_0` is the initial height in meters and t is a time in seconds. `H_0` is measured off the crater's bottom here.

We know that after the time `t_1=2.5` s the height becomes zero. So  H_0=g_M*(t`_1)^2/2 approx 11.9` (m).

On Earth this distance would be about 2.5 times greater due to greater gravity acceleration.

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