Question

A body dropped from top of a tower fall through $40\text{m}$ during the last two seconds of its fall. The height of the tower is
($g=10{\text{m/s}}^2$)

• A. $90\text{m}$
• B. $45\text{m}$
• C. $55\text{m}$
• D. $60\text{m}$

Answer 1

The correct option is B $45\text{m}$

Let the body fall through the height of the tower in $t$ seconds.
From equation of motion we have
$s_n=u+\frac{a}{2}(2n-1)$
So, we have total distance travelled in last $2$ seconds of fall as
$40=s_t+s_{(t-1)}$
$\Rightarrow 40=[0+\frac{g}{2}(2t-1\left)\right]+[0+\frac{g}{2}(2t-3\left)\right]$
$\Rightarrow 40=\frac{g}{2}(4t-4)\Rightarrow t=3\text{sec}$
Thus, distance travelled in $3\text{sec}$ will be the height of the tower which is given by
$h=ut+\frac{1}{2}g{t}^2=0+\frac{1}{2}\times 10\times 3^2$
$\Rightarrow h=45\text{m}$