A body is dropped from certain height H. If the ratio of the distances traveled by it in (n-3) seconds to (n-3)rd second is 4:3,find H.
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s=ut + 1/2 at^2
u=0
distance traveled in (n-3) sec
s=(1/2)g(n-3)^2
also
Sn= u+ (g/2)*(2n-1)
here n= (n-3)rd
and u=0
therefore, Sn= (g/2)* (2(n-3)-1)
Sn= (g/2) * (2n- 7)
so
It is given that S/Sn= 4/3
So,
[(1/2)g(n-3)^2] / [(g/2)*(2n-7)]=4/3
now calculate the value of n.
once calculated n, put it equal to t and find the value of h from h=1/2 gt^2
u=0
distance traveled in (n-3) sec
s=(1/2)g(n-3)^2
also
Sn= u+ (g/2)*(2n-1)
here n= (n-3)rd
and u=0
therefore, Sn= (g/2)* (2(n-3)-1)
Sn= (g/2) * (2n- 7)
so
It is given that S/Sn= 4/3
So,
[(1/2)g(n-3)^2] / [(g/2)*(2n-7)]=4/3
now calculate the value of n.
once calculated n, put it equal to t and find the value of h from h=1/2 gt^2
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