# When a ball falls from a height of 125 m, what is the distance traveled by the ball in the last second of its fall.

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### 1 Answer

A ball is dropped from a height of 125 m. The initial velocity of the ball is 0 m/s and the acceleration of the ball is a constant 9.8 m/s^2 in the downward direction.

If the time taken by the ball to drop 125 m is T. `(1/2)*9.8*T^2 = 125`

=> `T = sqrt(1250/49)` seconds.

The final velocity of the ball when it strikes the ground is `9.8*sqrt(1250/49)` . The velocity of the ball 1 second before it strikes the ground is `9.8*(sqrt(1250/49) - 1)`

The distance s traveled by the ball in the last second is given by `2*9.8*s = (9.8*sqrt(1250/49))^2 - (9.8*(sqrt(1250/49) - 1))^2`

=> `2*s = 9.8*(1250/49) - 9.8*(sqrt(1250/49) - 1)^2`

=> `s = 4.9*(1250/49 - 1250/49 - 1 + 2*sqrt(1250/49))`

=> `s = 4.9*(2*sqrt(1250/49) - 1)`

=> `s ~~ 44.59`

**The ball travels a distance equal to 44.59 m in the last second.**