Question

A body projected vertically up has displacement of $16m$ in the first 'n' seconds while it was moving up. Its magnitude of displacement in the last 'n' seconds while falling down is:

• A. $8m$
• B. $4m$
• C. $16m$
• D. $2m$

Answer 1

The correct option is C $16m$

The maximum vertical height or displacement of the project body upward is 16m in the first n seconds.

While going upwards, distance traveled in first 'n' seconds would be

$h_1=un-\frac{1}{2}g{n}^2...........\left(1\right)$

from $v=u-gt$ time taken to reach maximum height is

$T_{max}=u/g$

Total time of flight $=2T_{max}=2u/g...........\left(2\right)$

While going upwards, distance traveled in last 'n' seconds would be the difference of distance covered in time $2T_{max}$ and distance covered in time $2T_{max}-n$.

$h_2=S_{2T_{max}}-S_{2T_{max}-n}$

$=[u2T_{max}-\frac{1}{2}g(2T_{max}{)}^2]-[u(2T_{max}-n)-\frac{1}{2}g(2T_{max}-n{)}^2]$

$=un+g{n}^2/2-gn2T_{max}...............\left(3\right)$

From (2) and (3),

$h_2=-un+\frac{1}{2}g{n}^2...........\left(4\right)$

From (1) and (4), $h_1=-h_2$

$|h_2|=|h_1|$