Answer to Question #160864 in Classical Mechanics for Matthew Brousseau

Question

At a lab investigating fire extinguisher foams, a heavy ball is accidentally dropped into a deep vat of foam from a crane 5.30m above the foam. After entering the foam, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the foam. The ball reaches the bottom 3.60 s after it is released. How deep is the vat?

Answer 1

Let the deepness of the vat is h.

total covered distance by the ball = 5.30+h

Velocity of the ball after the fall of 5.3 m is $v^2= u^2+2g\times 5.3$

here u=0 for ball and $g =10m/sec^2$

$v=\sqrt{2\times 10\times 5.3} =\sqrt{106}m/s$

now, time taken by the ball to reach 5.3m is $(t_1)= \frac{v}{g}=\frac{\sqrt{106}}{10}$

let time taken by the ball to reach at the bottom, from the top of the vat is $t_2$

$3.6=t_1+t_2$

$\Rightarrow 3.6 = \frac{\sqrt{106}}{10}+\frac{h}{\sqrt{106}}$

$\Rightarrow \frac{h}{\sqrt{106}}=3.6-1.03$

$\Rightarrow \frac{h}{10.3}=2.57m$

$\Rightarrow h = 2.57\times 10.3m$

$\Rightarrow h =26.47m$

Hence, depth of the vat is 26.47 m