Answer:
17.3 m/s
Step-by-step explanation:
We can use the conservation of energy principle to find the velocity of the log when it hits the water. At the top of the waterfall, the log has gravitational potential energy, and at the bottom, it has kinetic energy. We can assume that there is no energy lost to friction or air resistance.
The gravitational potential energy of the log at the top of the waterfall is given by:
U = mgh
where m is the mass of the log, g is the acceleration due to gravity, and h is the height of the waterfall. Substituting the given values, we get:
U = (15.0 kg)(9.81 m/s^2)(30 m) = 4414.5 J
At the bottom of the waterfall, all of the gravitational potential energy has been converted to kinetic energy:
U = (1/2)mv^2
where v is the velocity of the log when it hits the water. Solving for v, we get:
v = sqrt(2U/m)
Substituting the value of U and m, we get:
v = sqrt(2(4414.5 J)/(15.0 kg)) = 17.3 m/s
Therefore, the velocity of the log when it hits the water is approximately 17.3 m/s.
Hope this helps!