An open elevator is ascending with zero acceleration the speed v = 10m/s a ball is thrown vet
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You have two frames of reference here.
As in your first part you want height above the earth then
we MUST use the earth frame of reference to solve it.
From the point at which the ball is released,
its velocity relative to the earth is 30 m/s
and from this point it gains a height given by s
= v^2 / 2g
= 900/( 2 * 9.8) = 46 m
so the final height from the earth is 30 + 46 = 76 m
Now for the second one we do not need to use the earth.
The ball is thrown from the elevator and falls back to the elevator.
From the elevator it will rise a distance of v^2 / 2g but v is only 20 m/s
So it rises 20.4 m
It must fall 22.4 m
Now as s = 1/2 g t^2
-> t = sqrt( 2s/g)
we can apply that to the two motions both up and down
t1+t2 = sqrt(2*20.4 / 9.8 ) + sqrt(2*22.4 /9.8 )
=4.18 s