A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later. How high is the cliff and how far from the base of the cliff did the diver hit the water?
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The height of the cliff is 19.6m and the base of the cliff did the diver hit the water is 7.2m

The distance at the base of the cliff did that the diver hit the water is the horizontal distance. This is expressed as:

Given the following parameters

Speed v = 3.60m/s

Time t = 2.0s

Substitute into the formula to have:

The height of the cliff will be the vertical height derived according to the Newton's law of motion:

Hence the height of the cliff is 19.6m

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Explanation:

It is given that,

The horizontal speed of a cliff diver,

It reaches the water below 2.00 s later, t = 2 s

Let is the distance where the diver hit the water. It can be calculated as follows :

Let is the height of the cliff. It can be calculated using second equation of motion as follows :

So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.