Question

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

• A. $60m$
• B. $45m$
• C. $80m$
• D. $50m$

Answer 1

The correct option is B $45m$

Let the body fall through the height of tower in $t$ seconds.
We know that distance covered the body $\left(D_n\right)$ at a particular second $n$ is given by the relation
$D_n=u+\frac{a}{2}(2n-1)$
As the body is at rest initially
$\Rightarrow u=0$
$\therefore D_n=\frac{a}{2}(2n-1)$
Let $t$ be the time taken by the body to reach ground.
Distance travelled in last $2$ seconds $=40m$ (given)
Total distance travelled in last 2 seconds of fall can also be expressed as
$D=D_t+D_{(t-1)}$
$=[0+\frac{g}{2}(2t-1\left)\right]+[0+\frac{g}{2}\{2(t-1)-1\left\}\right]$
$=\frac{g}{2}(2t-1)+\frac{g}{2}(2t-3)=\frac{g}{2}(4t-4)$
$=\frac{10}{2}\times 4(t-1)$
$\Rightarrow 40=20(t-1)$
$t=2+1=3s$
Distance travelled in $t$ seconds is
$s=ut+\frac{1}{2}a{t}^2=0+\frac{1}{2}\times 10\times 3^2=45m$