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:

`S=un+1/2g*n^2`

Here, `u=0`

So, `S=1/2*gn^2` .................(i)

Similarly, let the total distance covered in `(n-1)` secs. is `S'` , then:

`S'=1/2*g(n-1)^2`  ..................(ii)

Hence, the distance covered in the last second is given by `S-S'` .

According to the given problem:

`S-S'=50`

Substituting the values of `S` and `S'` from equations (i) and (ii) we get:

`1/2*gn^2-1/2*g(n-1)^2=50`

`rArr 1/2*g{n^2-(n-1)^2}=50`

`rArr g(2n-1)=100`

`rArr2n-1=100/g`

`rArr n=(100+g)/2g`

Plugging `g=9.81m/s^2` in the above equation we get:

`n~~ 5.6`

Putting `n=5.6` in equation (i),

`S=1/2*9.81*(5.6)^2=153.82 m`

Therefore, the height of the tower is approximately 153.82 m.

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