A plastic ball is dropped in water from a height of 45m. It goes in the water and then bounces back to the starting point. (Neglect viscosity). If the time taken in this process is 18seconds than the specific gravity of the material of plastic ball is
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The ball falls 45m in air and rises 45m in water. Neglecting viscosity, only gravity and buoyancy act. The net acceleration in water is a = g(1 - σ), where σ is specific gravity. Time in air: t_air = √(2h/g) = √(90/9.8) ≈ 3.03s. Total time is 18s, so time in water t_water = 18 - 2t_air ≈ 11.94s (up and down time equal). Using h = (1/2)at² for descent in water: 45 = (1/2)g(1-σ)(t_water/2)². Solving gives σ ≈ 0.75.
Final answer: 0.75