CBSE Class 11-science Answered

The velocity of the ball just before entering the water = (2x9.8x19.6)^1/2 = 19.6 m/s

Force of gravity = ρ_sV_sg

Buoyancy force = ρ_wV_sg

Net force = ρ_sV_sg - ρ_wV_sg = V_s(ρ_s-ρ_w)g = - 500x9.8xV_s

Net acceleration = Net force/mass = - 500x9.8xV_s/(500xV_s) = -9.8 m/s²

The time required to bring the ball to rest in water = 0 - 19.6 m/s / -9.8 m/s² = 2 sec

The depth it goes = 2(19.6) - 9.8x2x2/2 = 19.6 m.

It'll take 2+2 = 4 seconds to come again to the water surface after entry ino water.

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