Answer:
The value speed is [tex]v = 8.192 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the ping pong ball is [tex]m = 0.0027 \ kg[/tex]
The radius is [tex]r = 0.020 \ m[/tex]
The depth of the ping pong inside the water is [tex]s = 0.3 \ m[/tex]
Gnerally the force with which the ball will emerge will from the water is mathematically represented as
[tex]F = B - m g[/tex]
Here B is the Buoyant force on the ball which is mathematically represented as
[tex]B = \rho_w * V_b * g[/tex]
Here [tex]V_b[/tex] is the volume of the ball which is mathematically represented as
[tex]V_b = \frac{4}{3} * \pi * r^3[/tex]
=> [tex]V_b = \frac{4}{3} * 3.142 * 0.020 ^3[/tex]
=> [tex]V_b = 3.35*10^{-5} \ m^3[/tex]
[tex]\rho_w[/tex] is the density of water with value [tex]\rho_w = 1000 \ kg/m^3[/tex]
So
[tex]B = 1000 * 3.35*10^{-5} * 9.8[/tex]
=> [tex]B = 0.3284 \ N[/tex]
So
[tex]F = 0.3284 - 0.0027 * 9.8[/tex]
=> [tex]F = 0.3284 - 0.02646[/tex]
=> [tex]F = 0.3020 \ N[/tex]
Generally force is mathematically represented as
[tex]F = m * a[/tex]
So
[tex]0.3020 = 0.0027 * a[/tex]
=> [tex]a = 111.84 \ m/s^2[/tex]
Generally from kinematic equation
[tex]v^2 = u^2 + 2as[/tex]
Here u is the velocity of the ping pong at depth of 0.3 m and the value is zero given that at that point there was no motion
So
[tex]v^2 = 0^2 + 2 * 111. 84 * 0.3[/tex]
=> [tex]v = 8.192 \ m/s[/tex]
