Answer:
18.66m
Explanation:
This was actually fun! Let set up the problem first: As you drop the sphere, it will accelerate till it hits the water with a given speed [tex]\dot z[/tex]. Once sinking the ball is subjected to two forces: its own weight [tex]m_sg = \rho_sVg[/tex] directed towards the bottom of the lake, and buoyancy (Archimedes law), ie a force upward equal to the volume of displaced water, [tex]m_wg=\rho_wVg[/tex]. At this point it's the classical "how high can I toss a ball before it falls down" with a water twist. Please note that I'm using z as the height of the sphere, and [tex]\dot z; \ddot z[/tex] represent the velocity (with one dot) and the acceleration (with two dots) in the z direction .Let's assume the sign of all quantities being positive if directed upwards, and negative if towards the bottom of the lake.
Let's first determine how fast the ball hit the water. For me the easiest way is saying "at 8m it has a given potential energy, and 0 kinetic energy. When it hits the water it loses all potential energy and has only kinetic energy". In numbers:
[tex]m_sgz=\frac12m_s\dot z_0^2[/tex] Let's divide by the masses, g is a known value ([tex]9.81 ms^{-2}[/tex]), z is 8 meters, and we get that [tex]\dot z_0 = 4\sqrt g =12.52 m/s[/tex]
At this point, let's determine the force acting on the sphere, courtesy of Newton second law and the debate we had earlier:
[tex]m_s \ddot z = -m_sg + m_wg\\\rho_sV \ddot z = -\rho_sVg + \rho_wV g[/tex]
At this point we can divide by the volume of the sphere, and make use of the fact that [tex]\rho_s = 0.7 \rho_w[/tex]
[tex]0.7\rho_w \ddot z = -.7\rho_w g+ \rho_wg \rightarrow 0.7 \ddot z = 0.3 g\\\ddot z = \frac {0.3}{0.7} g= \frac 37g = 4.2 ms^-2[/tex]
We're almost there. At this point we can write the law of motion for the sphere.
[tex]\dot z = \dot z_0 + \ddot z t[/tex] that we will use to find the time for the ball to stop sinking - ie reaches 0 velocity.
[tex]0 = -12.52 +4.2t \rightarrow t= 12.52/4.2 \approx 3s[/tex]
At this point we can use the other law of motion to find out the distance traveled
[tex]z= z_0 + \dot z_ot + \frac12 \ddot z t^2\\z= 0 -12.52 (3) + \frac12 (4.2)(3)^2 = -37.56+18.90 = -18.66 m[/tex]
as usual, please double and triple check all the calculations, it's almost 1.30 am here and I am not the most confident number cruncher even when fully awake.
