A body is falling freely under gravity. How much distance it falls during an interval of time between 1st and 2nd seconds of its motion, taking g=10 ?

Using the kinematics equation, s = u.t + 1/2.a.t².
In t = 1sec, s = 1/2 (10) (1)² = 5m
In t = 2sec, s = 1/2 (10) (2)² = 20m
The distance travelled between the first and second seconds = 20 – 5 = 15m

how

How this answer is 14?which is suitable formula for this question

Hello this question answer is 15m
S1= 1/2gt^2 = 1/2(10)(1)^2 = 5m and
S2 = 1/2gt^2 = 1/2(10)(2)^2 = 1/2(10)(4) = 20m

so S = S2-S1
S =20 – 5 = 15m

D1=VT +1/2gt². At 2 second
D1= 0+1/2*10*4
D1= 40/2 =20m

At one second
D2=VT +1/2gt².
D2=0+1/2*10*1
D2= 5 m

D=D2-D1

D=20-5 =15 m

answer is 15 m
S=ut+1/2gt^2
u=0 ,g = 10ms-2
During ist time interval
S1=0+1/2(10)(1)^2
S1=5m
During 2nd time interval
S2=0+1/2(10)(2)^2
S2=20m
Distance covered by body during an interval of time between first and second seconds of its motion = (S2 – S1 ) = 20–5 = 15m . Hope you like it!

Explanation would have been helpful….

The answer is 20 m, kindly correct this