Question

A soild ball of density half that of water falls freely under gravity from a height of 19.6 m and then enters water. Upto what depth will the ball go. How much time will it take to come again to the water surface? Neglect air resistandce and viscosity effects in water. (Take g=9.8 m//S^(2)).

Answer 1

Given:

• Density of solid ball = 0.5 * Density of water
• Solid ball falls freely under gravity from H = 19.6 m
• It enters the water.

To find:

• Depth up to which ball will go = d.
• Time taken by ball to come again to water surface.

Answer:

• Taking mass of the ball = m and density of the ball = ρ .Therefore, density of water = 2ρ.

• Ball falls from a height of 19.6 m,

          Potential energy when ball reaches the surface = mgh = (mg×19.6) J

• As the ball strikes the water surface, upthrust acts. But weight drags the ball into water.

       Force opposing the motion = Upthrust - Weigh

                                                     = V (2ρ) g - mg    

                                                     = (m/ρ) (2ρ) g - mg

                                                     = mg

• Using work - energy theorem, we get:  

            Force opposing * Depth = Potential Energy

          ⇒ mg * d = mg * 19.6

          ⇒ d = Depth up to which ball will go = 19.6 m

• Velocity at water surface initially = v = $\sqrt{2 g H}$ = $\sqrt{2 * 9.8 * 19.6} = 19.6 m/s$

• u = 19.6 m/s , v = 0 (when it reaches the maximum depth)

           v = u + at

      ⇒ 0 = 19.6 - 9.8 t

      ⇒ t = 2 sec = Time from water surface to maximum depth

• Time to go to water surface again = 2 + 2 = 4 sec