Question

a body falls freely from the top of a towerand during the last second of its flight ,it fall 16/25th of the whole distance. find height of the tower.taje g=9.8m/s^2

Answer 1

Let the height of the tower be ‘h’. Suppose it takes ‘n’ second to reach the ground. Thus, in nthsecond the body covers a height (16h/25).
It starts from rest, thus,
Sn = u + (a/2)(2n – 1) [this is the equation to find the distance traveled in nth 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 gn2
=> h = (9.8/2)n2
=> h = 4.9n2 …………….…(2)
From (1) and (2) we have,
16(4.9n2)/25 = 4.9(2n – 1)
=> 16n2 = 25(2n – 1)
=> 16n2 – 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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