A body is dropped from the top of a tower. During the last second of its fall it covers 16/25th of the height of the tower. Calculate height of tower? I found a link answering this question but it wasn't satisfactory.Can you please explain it properly to me as in the link i didnt understand how 2n-1 came and the second part too.can u answer it again?? |

Let the height of the tower be ‘h’. Suppose it takes ‘n’ second to reach the ground. Thus, in n^{th} second the body covers a height (16h/25).

It starts from rest, thus,

S_{n} = u + (a/2)(2n – 1) [this is the equation to find the distance traveled in n^{th} second]

=> (16h/25) = 0 + (9.8/2)(2n – 1)

=> 16h/25 = 4.9(2n – 1) …...(1)

Again, the body reaches the ground in ‘n’ second. So, the distance traveled in ‘n’ second is,

h = 0 + ½ gn^{2}

=> h = (9.8/2)n^{2}

=> h = 4.9n^{2} …………….…(2)

From (1) and (2) we have,

16(4.9n^{2})/25 = 4.9(2n – 1)

=> 16n^{2} = 25(2n – 1)

=> 16n^{2} – 50n + 25 = 0

=> n = 0.625 or 2.5

Here, n = 0.625 s is not possible. So, the time in which the body hits the ground is 2.5 s.

So, (1) => h = 30.625 m

This is the height of the tower.

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