Q. A ball is dropped from rest from the top of a tower.The distance covered in the last second of its motion is 50 m. Find the height of the tower.
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Since, the ball is dropped from rest from the top of a tower, the total distance travelled by the ball is equal to the height of the tower.
Let the total distance `S` is covered in `n` seconds, then:
So, `S=1/2*gn^2` .................(i)
Similarly, let the total distance covered in `(n-1)` secs. is `S'` , then:
Hence, the distance covered in the last second is given by `S-S'` .
According to the given problem:
Substituting the values of `S` and `S'` from equations (i) and (ii) we get:
Plugging `g=9.81m/s^2` in the above equation we get:
Putting `n=5.6` in equation (i),
Therefore, the height of the tower is approximately 153.82 m.
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