Question

A body fahs 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 : 4 : 9
• B. 1 : 2 : 4
• C. 1 : 3 : 5
• D. 1:2:3

Answer 1

Answer: (C)

Using the relation
$s=ut+\frac{1}{2}g{t}^2$
As the body is falling from rest, u = 0$<br/>s=\frac{1}{2}g{t}^2$
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