Final answer:
The stone strikes the ground with a speed of approximately 10.84 m/s.
Step-by-step explanation:
The subject of this question is Kinematics. Kinematics is the branch of physics that deals with the motion of objects without considering the forces causing the motion.
To determine the speed with which the stone strikes the ground, we can use the Principle of Conservation of Energy. We know that the stone is dropped from a height of 6 m above the ground, and its mass is 25 g. The potential energy of the stone at 6 m above the ground is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. PE = (0.025 kg) * (9.8 m/s^2) * 6 m = 1.47 J.
The total mechanical energy of the stone is the sum of its potential energy and kinetic energy. In this case, since the stone is dropped, its initial kinetic energy is zero. Therefore, the total mechanical energy is equal to the potential energy: 1.47 J.
Using the equation KE = 0.5mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity, we can solve for the velocity. Rearranging the equation, we have v = sqrt(2 * KE / m) = sqrt(2 * 1.47 J / 0.025 kg) = sqrt(117.6) m/s = 10.84 m/s.
Therefore, the stone strikes the ground with a speed of approximately 10.84 m/s.