Question

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 the height of the tower.

Answer 1

Step 1: Given

Let the height of the tower be ‘h’.

Suppose it takes ‘n’ second to reach the ground.

'Sn' is height covered in $(n-1)$ second.

Step 2: Formula used

$S=ut+\frac{a}{2}{\left(t\right)}^2$

Step 3: Finding the time required to cover the height of the tower

In n^th second the body covers a height of $\frac{16}{25}$h.

In (n-1)th second, the body covers a height of (1-$\frac{16}{25}$h) = $\frac{9}{25}$h

It starts from rest, thus,

$Sn=u+\frac{a}{2}{(n-1)}^2$ [this is the equation to find the distance traveled in (n-1)th second]

= (9h/25) = 0 + (9.8/2)(n – 1)²

= 9h/25 = 4.9(n – 1)² …...(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,

$\frac{9\times 4.9\times n^2}{25}$ = 4.9(n – 1)²

=> 9n² = 25(n – 1)²

=> 16n² – 50n + 25 = 0

=> n = 0.625 or 2.5

Here, n = 0.625 s is not possible as it is greater than 1. So, the time in which the body hits the ground is 2.5 s.

Step 4: Finding the height of the tower

Putting value of n in equation (2), we get

h=4.9n2=4.9×2.5×2.5=30.625h = 4.9n^2=4.9\times 2.5\times 2.5=30.625h=4.9n2=4.9×2.5×2.5=30.625 m.

Hence, the height of the tower is 30.625 m.