A ball rolls horizontally off the edge of a cliff at 4.00 m/s. If the ball lands a distance of 30.0 m from the base of the vertical cliff , what is the height of the cliff?  

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When the ball rolls off the clip at the very first moment it has the horizontal velocity component which is 4m/s. If we assume that there is no air resistance the this horizontal velocity component remains same through out. So the ball lands away from the bottom of clip.

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When the ball rolls off the clip at the very first moment it has the horizontal velocity component which is 4m/s. If we assume that there is no air resistance the this horizontal velocity component remains same through out. So the ball lands away from the bottom of clip.

When the ball rolls off the cliff due to its weight it will come downwards with gravitational acceleration.

Using euqations of motion;

`rarrs = ut+1/2at^2`

At the horizontal direction we have no acceleration.

`30 = 4xxt+0`

`t = 30/4`

`t = 7.5s`

`darrs = ut+1/2at^2`

Since the ball rolls horizontally at the edge of the clip we don't have a vertical component.

`h = 0+1/2xx9.81xx7.5^2`

`h = 275.9m `

So the height of the clip is 275.9m

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