What is the distance covered by a freely falling body during the first seconds of it's motion ( g = 10 m/s)
Answers
Answered by
42
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
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
Answered by
25
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✌
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✌
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