an object takes 5s to reach the ground from a height of 5m on a planet what is the value of g on the planet
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s=ut + (1/2)at^2
s= 0 + 0.5 * a * 5^2 ( assuming it as a free drop... so u=0 )
s= 12.5a
12.5a=5
a=5/12.5
a=0.4m/s ( a and g both are same)
s= 0 + 0.5 * a * 5^2 ( assuming it as a free drop... so u=0 )
s= 12.5a
12.5a=5
a=5/12.5
a=0.4m/s ( a and g both are same)
Answered by
4
HEY USER ,
HERE IS YOUR ANSWER BuDDY :-
As we know :-
s = ut + 1/2at^2
let h be the distance and g is the acceleration put these values in formula we will get ,
h = ut + 1/2 gt^2
as the initial velocity i.e. u = 0
then , g = 2h / t^2
now the value of h is 5m and for time t is 5sec .
So , g= 2×5/ 5^2
g= 0.4m/sec^2 .....
HOPE It HELPS YOU !!!!!
mark me as the brainliest if helps....plzz
HERE IS YOUR ANSWER BuDDY :-
As we know :-
s = ut + 1/2at^2
let h be the distance and g is the acceleration put these values in formula we will get ,
h = ut + 1/2 gt^2
as the initial velocity i.e. u = 0
then , g = 2h / t^2
now the value of h is 5m and for time t is 5sec .
So , g= 2×5/ 5^2
g= 0.4m/sec^2 .....
HOPE It HELPS YOU !!!!!
mark me as the brainliest if helps....plzz
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