Question

The distance travelled by a body falling freely from rest in 2nd, 3rd, and 5th second of its motion are in the ratio
A) 7:5:3 B) 3:5:9
C) 5:3:7 D) 5:7:3

Answer 1

Dear student,
Distance travelled in 'n'th second-
S=u+a/2×{2n-1} where, u=0, a=g=10m/s²​​​​​​
​​​distance travelled in fifth second-
S₅=0+10/2×{(2x5)-1}
S₅=45
distance travelled in second second-
S₂=0+10/2×{(2×2)-1}
S₂=15
distance travelled in third second-
S₃=0+10/2×{(2×3)-1}
S₃=25
So,
S₂:S₃:S₅=15:25:45=3:5:9
Regards