A stone drop from height h reaches at earth surface in 1 sec. If the same stone taken to moon and drop freely from height h then it will reaches at the surface of moon in the time?
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On earth h=(1÷2)gt×th=5on moon,a=g÷65=(1÷2)(g÷6)t×t6=t×tt=√6 if you got it reply me
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We have,
S=ut+a(t squared)/2
On earth S=h
U=0
T=1sec
a=10m/s2
h=0(1)+10(1 square)/2
h=5 metres
On moon acceleration=1/6 of acceleration on earth
A=10/6
S=h=5m
U= 0
t=?
S=ut+at2/2
5=0+(10/6)*(t squared)÷2
1=t squared/6
=>t squared=6
=>[t=square root of 6]
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