A ball of mass m=0.25 kg is dropped from rest. What is the speed of the ball after it falls two meters?
V^2 = Vo^2 + 2g*h = 0 + 19.6*2 = 39.2
V = 6.26 m/s.
To find the speed of the ball after it falls two meters, we can use the principles of physics, specifically the law of conservation of energy. The potential energy gained by the ball as it falls is converted into kinetic energy, which is related to the speed.
1. First, let's calculate the potential energy gained by the ball as it falls two meters. The potential energy (PE) can be calculated using the formula PE = mgh, where m is the mass (0.25 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (2 meters).
PE = 0.25 kg * 9.8 m/s^2 * 2 meters
PE = 4.9 Joules
2. Now that we have the potential energy, we can equate it to the kinetic energy (KE) of the ball. The kinetic energy is given by the formula KE = 0.5mv^2, where v is the velocity (speed) of the ball.
KE = 0.5 * 0.25 kg * v^2
3. Since energy is conserved, we can equate the potential energy gained to the kinetic energy.
PE = KE
4.9 Joules = 0.5 * 0.25 kg * v^2
4. Now we can solve for the velocity (speed) of the ball. Rearrange the equation to solve for v:
v^2 = (2 * 4.9 Joules) / 0.25 kg
v^2 = 39.2 m^2/s^2
Taking the square root of both sides:
v = √(39.2 m^2/s^2)
v ≈ 6.26 m/s
Therefore, the speed of the ball after it falls two meters is approximately 6.26 m/s.