Question

What is the distance covered by a freely falling body during the first seconds of it's motion ( g = 10 m/s)

Answer 1

using 4th eqn. of motion
S(nth)=u+a/2(2n-1)
s is the distance
u is the intial velocity i.e. 0
a =10m/s
n is the no. of seconds
s=0+10/2(2*1-1)
s=10/2(2-1)
s=5*1=5m
the distance covered by the object In a free fall in 1st sec is 5m

Answer 2

To calculate the displacement covered by an object in first n seconds, this formula is used:-

S(n) = g/2(2n-1)

Put n = 1 then,

S(n) = 10/2[2(1)-1]
= 5 × 1 metres
= 5 metres

This formula is very useful, try to remember it...

Hope it helps✌