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 height of the tower ......​

Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\Large\bold{\boxed{\boxed{30.6\:m }}}$

$\rule{200}{4}$

$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

Let us assume ‘h’ be the height of the tower & the time of flight be ‘n’ seconds.

Here, initial velocity (u) is zero

The equation to find the distance traveled in nth second is

$\Large\bold{\boxed{\boxed{S_n= u+\frac{1}{2}a(2n-1)}}}$

$\bf\bold{\Rightarrow\frac{16}{25}h= 0+\frac{1}{2}9.8(2n-1)}$

$\bf\bold{\Rightarrow\frac{16}{122.5}h= (2n-1)}$

$\bf\bold{\Rightarrow\:n=\frac{1}{2}(\frac{16}{122.5}+1)}$ - - - - - -(1)

A/q

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

$\bf\bold{\Rightarrow\:h=ut+\frac{1}{2}gn^2}$

$\bf\bold{\Rightarrow\:h=0+\frac{1}{2}9.8\times[\frac{1}{2}(\frac{16}{122.5}h+1)]}$ - - - - (from eq 1)

On solving

=> h = 30.625 m

$\bf\bold{\Rightarrow\:h≈30.6 \: m}$

Hence, the height of the tower is 30.6 m

Answer 2

${\textbf{Answer}}$

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 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.