Question

A body starts falling from height 'h' and travels distance h/2 during the last second of motion. The time of travel (in sec.) is:
(A) √2 - 1
(B)2 + √2
(C) √2 + √3
(D) √3+2
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Answer 1

Answer

Option (B) is correct

Given

A body starts falling from height 'h' and travels distance h/2 during the last second of motion

To Find

The time of travel (in sec)

Solution

Case - 1 : For whole distance , s

Apply 2nd equation of motion ,

$\to \rm s=ut+\dfrac{1}{2}gt^2\\\\\to \rm s=\dfrac{1}{2}gt^2\ ...(1) [\; \because\ u=0\ ]$

Case - 2 : For half distance

$\to \rm \dfrac{s}{2}=ut+\dfrac{1}{2}g(t-1)^2\\\\\to \rm \dfrac{s}{2}=\dfrac{1}{2}gt^2\ ...(2)[\; \because\ u=0\ ]$

Solve (1)/(2) ,

$\to \rm \dfrac{s}{\frac{s}{2}}=\dfrac{\frac{1}{2}gt^2}{\frac{1}{2}g(t-1)^2}\\\\\to \rm 2=\dfrac{t^2}{(t-1)^2}\\\\\to \rm \dfrac{t}{t-1}=\sqrt{2}\\\\\to \rm t=\sqrt{2}(t-1)\\\\\to \rm t=\sqrt{2}t-\sqrt{2}\\\\\to \rm t(\sqrt{2}-1)=\sqrt{2}\\\\\to \rm t=\dfrac{\sqrt{2}}{\sqrt{2}-1}\\\\\to \rm t=\dfrac{\sqrt{2}}{{\sqrt{2}-1}}\times \dfrac{{\sqrt{2}+1}}{{\sqrt{2}+1}}\\\\\to \rm t=\dfrac{\sqrt{2}({\sqrt{2}+1})}{2-1}\\\\\to \bf t=2+\sqrt{2}$