Question

A body falling for 2 seconds covers a distance S equal to that covered in next second. Taking $g=10m/s^2,S=$

• A. 30 m
• B. 10 m
• C. 60 m
• D. 20 m

Answer 1

The correct option is A 30 m

If U is the initial velocity then distance covered by it 2 sec.
$S=ut+\frac{1}{2}a{t}^2=u\times 2+\frac{1}{2}\times 10\times 4=2u+20..,\left(i\right)$

Now distance covered by it in $3^{rd}$sec.
$S_{3^{rd}}=u+\frac{g}{2}(2\times 3-1)10=u+25....\left(ii\right)$

From(i) and (ii),2u+20=u+25$\Rightarrow$u=5
$\therefore$S=2 $\times$ 5+20=30 m.