A 0.125-kg baseball is dropped from rest.
If the magnitude of the baseball's momentum is 0.750 kg * m/s
just before it lands on the ground, from what height was it dropped?
To determine the height from which the baseball was dropped, we can use the principle of conservation of momentum. The momentum of an object is the product of its mass and velocity. Since the baseball is dropped from rest, its initial velocity is zero.
We are given that the magnitude of the baseball's momentum just before it lands on the ground is 0.750 kg * m/s. We can calculate the final velocity of the baseball using the formula:
momentum = mass * velocity (p = m * v)
0.750 kg * m/s = 0.125 kg * v
Solving for the final velocity, we find that v = 6 m/s.
Next, we can use the principles of motion to find the height from which the baseball was dropped. The equation that relates displacement, initial velocity, final velocity, and acceleration under constant acceleration is:
displacement = (final velocity^2 - initial velocity^2) / (2 * acceleration)
Since the baseball is dropped from rest, the initial velocity is 0 m/s. And since it falls under the acceleration due to gravity, we can use g = 9.8 m/s^2 as the acceleration.
Plugging in the values, we have:
displacement = (6 m/s)^2 / (2 * 9.8 m/s^2)
Simplifying the equation gives us:
displacement = 18 / 19.6
Finally, the displacement represents the change in height. Therefore, we conclude that the baseball was dropped from a height of approximately 0.918 meters.