Question

A body falls freely from the top of the tower and during the last second of its flight,it falls 16/25th of the whole distance.find height of tower.(g=10m/s^2).

Answer 1

Let the height of the tower be ‘h’. Suppose it takes ‘n’ second to reach the ground. Thus, in nth second 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 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.

Answer 2

Answer:

125 m

Explanation:

Let H be the height of the building and t be the time to fall through this height H.

Then H = ½ g t^2 = 5 t^2. ----------------- (1)

In (t-1) second the distance traveled is 1 - 0.36 H = 0.64 H.

0.64 H = 5 (t-1)^2.---------------------------(2)

(2) / (1) gives 0.64 = (t-1)^2/ t^2

Or 0.8 = (t-1) / t

Solving we get t= 5 second.

Using (1) we get H = 125 m.