Question

A body is dropped from the top of a tower covers $\frac{7}{16}$ of total height isn the last second of its fall. The time of fall is

• A. $2s$
• B. $4s$
• C. $<1s$
• D. $>5s$

Answer 1

The correct option is B $4s$

Let us take the total time of fall be 't' sec.

Since, the body is dropped, its initial velocity$\left(u\right)=0$

So,

the total height(H) will be

$H=\frac{1}{2}g{t}^2$

at $(t-1)$ second

$x=\frac{1}{2}g(t-1{)}^2$

$\Rightarrow$ Total distance covered in the last second$=H-x$

So, according to question,

$H-x=\frac{3}{8}H$

$\frac{1}{2}g{t}^2-\frac{1}{2}g(t-1{)}^2=\frac{7}{16}\left[\frac{1}{2}g{t}^2\right]$

$\therefore (t>1)$

$t^2-(t-1{)}^2=\frac{7t^2}{16}$

$(t-1{)}^2=\frac{9t^2}{16}$

$\Rightarrow t-1=\pm \frac{3t}{4}$

$\Rightarrow t-1=\frac{3t}{4}$

$\Rightarrow t=48$ or $t-1=\frac{-3t}{4}$

$\Rightarrow t=\frac{4}{7}$s

Since, $t>1r$

so, total time of fall$=4$sec.