If a rock has a speed of 12 m/s as it hits the ground, from what height did it fall, assuming its initial speed was zero? Its mass is 50 kg.
mass does not matter.
s = 1/2 at^2
v = at, so t = v/a
s = 1/2 a (v/a)^2 = v^2/2a = 144/19.6 = 7.3m
To determine the height from which the rock fell, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system, consisting of the rock and the Earth, remains constant throughout the motion.
The total mechanical energy (E) is given by the sum of the kinetic energy (KE) and the gravitational potential energy (PE):
E = KE + PE
Initially, when the rock is at rest, the kinetic energy is zero. Therefore, only the gravitational potential energy is present. The gravitational potential energy (PE) can be calculated using the formula:
PE = m * g * h
where m is the mass of the rock, g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth), and h is the height from which the rock falls.
When the rock hits the ground, it has a speed of 12 m/s. At this point, all of its initial potential energy has been converted to kinetic energy. Using the formula for kinetic energy:
KE = (1/2) * m * v^2
where v is the final velocity of the rock, we can set up an equation by equating the initial potential energy to the final kinetic energy:
m * g * h = (1/2) * m * v^2
Simplifying and rearranging the equation:
h = (1/2) * v^2 / g
Substituting the given values:
h = (1/2) * 12^2 / 9.8
Calculating:
h ≈ 7.35 meters
Therefore, the rock fell from a height of approximately 7.35 meters.