A 0.135 kg ball is dropped from rest. If the magnitude of the ball's momentum is 0.730 kg·m/s just before it lands on the ground, from what height was it dropped?

this is a repeat of the question before it.

find its velocity from the momentum.

then vf^2=2gh find h.

jkj

To answer this question, we can use the principle of conservation of mechanical energy. The total mechanical energy of the ball is the sum of its kinetic energy and potential energy.

The ball is initially at rest, so it has no kinetic energy. Therefore, its initial mechanical energy is equal to its initial potential energy.

The final mechanical energy of the ball, just before it lands, consists only of its kinetic energy since its potential energy becomes zero at ground level.

Using the given information, we know that the magnitude of the ball's momentum is 0.730 kg·m/s just before it lands. We can find its final kinetic energy using the formula:

Kinetic Energy = (1/2) * mass * velocity^2

We can rearrange the equation to solve for velocity:

velocity = sqrt((2 * Kinetic Energy) / mass)

Substituting the given values:

velocity = sqrt((2 * 0.730 kg·m/s) / 0.135 kg)

velocity ≈ sqrt(10.8148 m^2/s^2) ≈ 3.29 m/s

Now, we can calculate the final mechanical energy (which only consists of kinetic energy):

Final Mechanical Energy = (1/2) * mass * velocity^2

Final Mechanical Energy = (1/2) * 0.135 kg * (3.29 m/s)^2

Final Mechanical Energy ≈ 0.284 J

Since the total mechanical energy is conserved, this final mechanical energy is also equal to the initial potential energy:

Initial Potential Energy ≈ 0.284 J

The potential energy can be calculated using the formula:

Potential Energy = mass * gravity * height

We can rearrange the equation to solve for height:

height = Potential Energy / (mass * gravity)

Substituting the known values:

height = 0.284 J / (0.135 kg * 9.8 m/s^2)

height ≈ 1.98 m

Therefore, the ball was dropped from a height of approximately 1.98 meters.