A large piece of wood, of mass 56 kg, falls over a waterfall and into the drop pool below. It enters the pool travelling at 12 ms-1 and is then subject to an upwards acceleration of 5.8 ms-2 due to its buoyancy. How long does it take before it bobs to the surface of the pool?

To solve this problem, we can use the equation that relates the distance traveled, initial velocity, acceleration, and time:

distance = initial velocity * time + 0.5 * acceleration * time^2

In this case, we want to find the time it takes for the wood to bob to the surface of the pool, so the distance we can use is zero (since it starts from the bottom of the pool). The initial velocity is 12 m/s and the acceleration is 5.8 m/s^2.

Plugging in these values into the equation:

0 = 12 * t + 0.5 * 5.8 * t^2

Simplifying the equation:

0.5 * 5.8 * t^2 + 12 * t = 0

This is a quadratic equation that we can solve for t using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = 0.5 * 5.8, b = 12, and c = 0. Plugging in these values:

t = (-12 ± √(12^2 - 4 * 0.5 * 5.8 * 0)) / (2 * 0.5 * 5.8)

Simplifying further:

t = (-12 ± √(144)) / 5.8

t = (-12 ± 12) / 5.8

Now we have two solutions:

t1 = 0 (since time cannot be negative)
t2 = (-12 + 12) / 5.8 = 0

Therefore, it takes 0 seconds for the wood to bob to the surface of the pool.