Question

A body dropped freely has covered $(16/25{)}^{th}$ of the total distance in the last second. Its total time of fall is

• A. $2.5s$
• B. $5s$
• C. $7.5s$
• D. $1s$

Answer 1

The correct option is A $2.5s$

Let total time $um/s$
So, distance travelled in last sec:
$\Rightarrow s_n=\left[u\right(u)+\frac{1}{2}g{n}^2]-\left[u\right(n-1)+\frac{1}{2}g(n-1{)}^2]$
$\Rightarrow s_n=\frac{1}{2}g(n^2-(n^2-2n+1\left)\right)=\frac{1}{2}g(2n-1)$

Distance total $\Rightarrow s=\frac{1}{2}g{n}^2$
$\Rightarrow \frac{16}{25}\left(\frac{1}{2}g{n}^2\right)=\frac{1}{2}g(2n-1)$

$\Rightarrow \frac{16}{25}{n}^2=2n-1$

$\Rightarrow 16n^2-50n+25=0$

$n=\frac{50\pm \sqrt{{50}^2-4\times 25\times 16}}{32}=\frac{50\pm \sqrt{900}}{35}$

$=\frac{50\pm 30}{32}=\frac{80}{32},\frac{20}{32}$

$n=2.5,0.625$
$\Rightarrow n=2.5sec$