Question

an elevator without a ceiling is ascending with a constant speed of 10m/s. a boy on the elevator shoots a ball directly upward, from a height of 2.0 m above the elevator floor. At this time the elevator floor is 28 m above the ground. the initial speed of the ball with respect to the elevator is 20 m/s (take g=9.8m/s²)
a). what maximum height above the ground does the ball reach?
b). How long does it take to return to the elevator floor

Answer 1

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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

Answer 2

Answer:

Hope it will help you surely......