Question

The distance travelled by a body falling freely from rest in 1st, 2 and 3 seconds are in the ratio

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