Respuesta :
First, we need to find the speed of the ball using the equation of motion: [tex] \sqrt{2gh} [/tex] with g = 9.8 and h=2
[tex] \sqrt{2*9.8*2} [/tex] = 6.32 m/s
Now we need to calculate the force which the water used to try to stop the ball using the Buoyancy force d*v*g
d*v*g = ma
a = [tex] \frac{dvg}{m} [/tex]
Let the density of the ball be d'
m=d'V
a= [tex] \frac{dvg}{d'v} [/tex]
a= [tex] \frac{dg}{d'} [/tex]
a= [tex] \frac{d}{d'} [/tex] * g
Relative density d/d' = 0.8
a = [tex] \frac{g}{0.8} [/tex]
a= [tex] \frac{9.8}{0.8} [/tex]
a = 12.25 [tex] m^{2} [/tex]/s
 deceleration a'=a-g = 2.45 [tex] m^{2} [/tex]/s
The final speed of the ball is v'=0
-v'^2 = v^2 + 2a's
40 = 0+2*2.5s
s=8m
The depth to which the ball will sink is 8m
Hope this Helps! :D
[tex] \sqrt{2*9.8*2} [/tex] = 6.32 m/s
Now we need to calculate the force which the water used to try to stop the ball using the Buoyancy force d*v*g
d*v*g = ma
a = [tex] \frac{dvg}{m} [/tex]
Let the density of the ball be d'
m=d'V
a= [tex] \frac{dvg}{d'v} [/tex]
a= [tex] \frac{dg}{d'} [/tex]
a= [tex] \frac{d}{d'} [/tex] * g
Relative density d/d' = 0.8
a = [tex] \frac{g}{0.8} [/tex]
a= [tex] \frac{9.8}{0.8} [/tex]
a = 12.25 [tex] m^{2} [/tex]/s
 deceleration a'=a-g = 2.45 [tex] m^{2} [/tex]/s
The final speed of the ball is v'=0
-v'^2 = v^2 + 2a's
40 = 0+2*2.5s
s=8m
The depth to which the ball will sink is 8m
Hope this Helps! :D
