What is the time taken by a stone dropped from a height of 245 m to reach the bottom?

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The distance (s) traveled by an object starting with an initial velocity u in time t is given by the formula s = u*t + (1/2)*a*t^2 where a is the average acceleration of the object.

When a stone is dropped from a height 245 m, its initial velocity is equal to 0 m/s. The acceleration of the stone is equal to the acceleration due to the gravitational force of attraction of the Earth that is 9.8 m/s^2. Substituting the values from the problem in the formula provided earlier gives: 245 = 0*t + (1/2)*9.8*t^2

`t^2 = (245*2)/9.8`

=> `t = sqrt(490/9.8)`

=> `t = sqrt 50`

=> t = 7.07

The stone takes approximately 7.07 seconds to reach the bottom when it is dropped from a height of 245 m.

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