Question

If a freely falling body travels in the last second, a distance equal to the distance travelled by it in the first three second, the time of the travel is

• A. $6sec$
• B. $4sec$
• C. $3sec$
• D. $5sec$

Answer 1

The correct option is D $5sec$

Given, body is freely falling.
So, its initial velocity, $u=0$
Distance travelled in first three second,
$s=ut+\frac{1}{2}a{t}^2$

$s_{\text{Three}}=0+\frac{1}{2}\times 10\times 9=45m$

Let the total time in which body is in motion be $t$.
The distance travelled in last second.
$s_n=u+\frac{a}{2}(2n-1)$

$s_{\text{Last}}=0+\frac{g}{2}(2t-1)$

And according to problem $s_{\text{Last}}=s_{\text{Three}}$
$45=\frac{1}{2}\times 10(2t-1)$

$⟹5(2t-1)=45$

$⟹2t-1=9$

$⟹t=5sec$