Question

A body falls from a height h = 200 m. The ratio of distance travelled in each 2 s, during t = 0 to t = 6s of the journey is

• A. 1:04:09
• B. 1:02:04
• C. 1:03:05
• D. 0.043090278

Answer 1

Answer: (C)

1:03:05

Answer 2

Using the relation
$s=ut+\frac{1}{2}gt^2$
As the body is falling from rest, u = 0$s=\frac{1}{2}gt^2$
Suppose the distance travelled in t = 2s, t = 4s, t = 6s are $s_2, s_4 \,and\, s_6$ respectively.
now
$s_2=\frac{1}{2}g(2)^2=2g$
$s_4=\frac{1}{2}g(4)^2=8g$
$s_6=\frac{1}{2}g(6)^2=18g$
Hence, the distance travelled in first two seconds
$(s_i)_2=s_2-s_0=2g$
$(s_m )_2 = ,s_4 -s_2=8g-2g =6g$
$(s_f)_2=s_6-s_4=18g-8g = 10g$
Now, the ratio becomes the ratio becomes
= 2g:6g: 10g = 1:3:5