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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[h=ut+\frac{ 1 }{ 2 }g t^2\] for each initial velocity u=0 m/s
\[g ~t^2=2h,g=\frac{ 2h }{ t^2 }\]
h is height t=time taken g=acceleration due to gravity
@surjithayer What is the acceleration due to gravity for the Earth, Moon, and Mars?
would t= 6.6 for the first empty slot?
distance m=height=h Time T=t i solve first \[g=\frac{ 2*2.25 }{ \left( 0.680 \right)^2 }=?\]
9.7
9.732
d = 1/2at2 is the equation for an object that starts at rest.
ok so- that is the one for the first box?
what would I do for the second box going horizontal?
when the body falls from a certain height,initial velocity=0 m/s
So what how would I figure out acceleration due to gravity for the second box going across?
h=3.08,t=0.793
so I would use that to figure out the second box?? @surjithayer
first two values are for earth next two for Moon and last two for Mars.
so the very first box on the graph is 9.732 right???
correct
please help with the rest
Hm

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