Question

A body is dropped from certain height H. If the ratio of the distances travelled by it in (n - 3) seconds to $(n-3{)}^{rd}$ second is 4 : 3, find H. (Take g = 10 m $s^{-2}$)

• A. 75 m
• B. 100 m
• C. 125 m
• D. 150 m

Answer 1

The correct option is C 125 m

Given,

$g=10m/s^2$

$S_n=u+\frac{a(2n-1)}{2}......1$ Where $S_n$ is the distance covered in $n^{th}$ second,

So, distance covered in $(n-3{)}^{th}$ in equation 1,

$\frac{a(2n-7)}{2}=5(2n-7).........2$

Replacing $n$ by $(n-3)$ in equation 1 here $u=0,a=10$

Therefore,

Distance covered in $(n-3)s=\frac{1}{2}a{t}^2=5(n-3{)}^2....3$

A/c $\frac{5(n-3{)}^2}{5(2n-7)}=\frac{4}{3}\Rightarrow n=5s$

Thus, $\mu =\frac{1}{2}a{n}^2=\frac{1}{2}10\times 5^2=125m$