# A coin is thrown horizontally with a speed of 12m/s from the top of a bridge. If the coin strikes the sea water below the bridge after 3s , find the height of the bridge above sea level.

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We are given the initial speed of the coin and the time it took before it hit the water.

The given initial speed is the horizontal speed -- that is, the horizontal component of the initial speed. The horizontal component of the speed of an object in projectile motion remains constant. The horizontal component of the speed will only give the horizontal distance travelled, given the time (but we don't need it in this problem).

The formula for time is:

`t = sqrt((2h)/g)`

where h is the height, and g is the acceleration due to gravity, `9.8 m/s^2` .

We know the time, so we only need to solve for h.

`t^2 = (2h)/g`

`2h = g t^2`

`h = (g t^2)/2`

Substituting the given values:

`h = (9.8*3^2)/2`

`h = 44.1`

The bridge is 44.1 meters above sea level.

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