Question

A freely falling body, fall from $45m$, covers of total distance in $3rd$ second.

• A. $8$%
• B. $35.5$%
• C. $55.55$%
• D. $45.5$%

Answer 1

Answer: (C)

$55.55$%

Total height $h=45m$.

apply second equation of motion-

$h=ut+\frac{1}{2}a{t}^2$

$u=0$; freely falling so initial velocity is zero.

$a=10m{s}^{-2}$

$\Rightarrow 45=0+\frac{1}{2}\times 10\times t^2$

$\Rightarrow t^2=9$

$\Rightarrow t=3s$

Now, Distance covered in initial two second.

$h_2=ut+\frac{1}{2}a{t}^2$

$\Rightarrow h_2=0+\frac{1}{2}\times 10\times 2^2$

$\Rightarrow h_2=20m$

So, distnace in $3rd$ second-

$S^{3rd}=45-20$

$\Rightarrow S^{3rd}=25m$

Total percentage -

$S^{3rd}=\frac{25}{45}\times 100$

$\Rightarrow S^{3rd}=55.55$