Explain impulse and calculate the impulse experienced by a ball during a collision, and determine the initial height from which it was dropped.

Mass of the ball ($m$) = $0.15 \text{ kg}$.
Velocity just before collision ($v_{before}$) = $6.2 \text{ m/s}$ (downwards).
Velocity just after collision ($v_{after}$) = $3.62 \text{ m/s}$ (upwards).
Let's choose upwards as the positive direction.
So, $v_i = -6.2 \text{ m/s}$ (initial velocity for the collision, downwards).
And $v_f = +3.62 \text{ m/s}$ (final velocity for the collision, upwards).

Impulse ($J$) is the change in momentum: $J = \Delta p = m v_f - m v_i$.

The magnitude of the impulse is $\boxed{\text{1.47 kg}\cdot\text{m/s}}$.

4.3:

The ball is dropped from rest ($v_{initial\_drop} = 0 \text{ m/s}$) and strikes the floor at $v_{impact} = 6.2 \text{ m/s}$. We use the conservation of mechanical energy.
$PE_{initial} + KE_{initial} = PE_{impact} + KE_{impact}$
Let the floor be the reference point for potential energy ($PE_{impact} = 0$).

Using $g = 9.8 \text{ m/s}^2$:

The requirement is to bounce to one third of the height it was dropped from.

The problem states the ball bounces to a height of $0.65 \text{ m}$. So, $h_{actual} = 0.65 \text{ m}$.
Comparing $h_{actual}$ and $h_{required}$:
$0.65 \text{ m}$ vs $0.6537... \text{ m}$
Since $0.65 \text{ m} < 0.6537... \text{ m}$, the ball does not meet the minimum requirements.

The ball does not meet the minimum requirements because its actual bounce height of $\boxed{\text{0.65 m}}$ is less than the required bounce height of $\boxed{\text{0.654 m}}$.