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Suppose that you release a small ball from rest at a depth of 0.520 m below the surface in a pool of water. If the density of the ball is 0.285 that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball.)
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Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

Suppose that you release a small ball from rest at a depth of0.560 m below the surface in a pool ofwater. If the density of the ball is 0.260 that of water and if the drag force on theball from the water is negligible, how high above the water surfacewill the ball shoot as it emerges from the water? (Neglect anytransfer of energy to the splashing and waves produced by theemerging ball.)

m

Herenet upward force acting on the ball is F = Buoyant force -weight

Buyantforce = volume * density of water * g

weight= Volume * density of the ball * g

Then F = Volume*g ( density of water - density of ball)

             = Volume *g*density of ball ( density of water/density ofball - 1)

             = mg ( density of water/density of ball - 1)

m =mass of the ball

       densityof water/density of ball = 1/0.26

Then F= mg(1/0.26   - 1)

          = 2.846*mg

Now netforce ma =  2.846*mg

Thenacceleration a = 2.846 *g

Nowinitial velocity of the ball at the depth is u = 0

final velocityat the surface is v

depth h= 0.56 m

Thenapply v² -u² = 2ah

                 v= ?2*2.846*9.8 *0.56    m/s

                  = 5.589 m/s

Nowfrom the surface initial velocity of the ball is u ' = v= 5.589 m/s

Nowthere is no water then it moves under the influence ofgravity

Then at maximumheight from the surface of water is v' = 0

Thenapply v² -u² = -2gh'

Then h' = u² /2gh' = ( 5.589m/s)²/2*9.8m/s ² = 1.59376 m