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Let's go through the solution for each part of the problem.

### Given Data:
- Block mass \( m = 30 \, \text{kg} \)
- Distance to fall \( h = 2 \, \text{m} \)
- Pulley mass \( M = 5 \, \text{kg} \)
- Pulley radius \( R = 0.1 \, \text{m} \) (20 cm diameter)
- Pulley moment of inertia \( I = \frac{1}{2}MR^2 \)
- We assume the rope doesn't slip on the pulley.

---

### Solution Steps

#### Part (a): Velocity of the 30 kg block just before it hits the ground

Using energy conservation, the gravitational potential energy of the block will be converted into:
1. Translational kinetic energy of the block.
2. Rotational kinetic energy of the pulley.

1. **Gravitational potential energy (initial)**:
   \[
   U = mgh = 30 \times 9.8 \times 2 = 588 \, \text{J}
   \]

2. **Kinetic energy (final)**:
   At the point just before impact, the block has kinetic energy:
   \[
   KE_{\text{block}} = \frac{1}{2}mv^2
   \]
   The pulley has rotational kinetic energy:
   \[
   KE_{\text{pulley}} = \frac{1}{2}I\omega^2
   \]
   where \(\omega\) is the angular velocity of the pulley. Since \(v = \omega R\), we can substitute \(\omega = \frac{v}{R}\).

   The total energy conservation equation is:
   \[
   mgh = \frac{1}{2}mv^2 + \frac{1}{2} \left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2
   \]

3. **Simplify and solve for \( v \)**:
   Substitute values and solve for \(v\):
   \[
   588 = \frac{1}{2}(30)v^2 + \frac{1}{2} \cdot \frac{1}{2}(5)(0.1)^2 \cdot \frac{v^2}{(0.1)^2}
   \]
   \[
   588 = 15v^2 + \frac{1}{4} \cdot 5 \cdot v^2
   \]
   \[
   588 = 15v^2 + 1.25v^2
   \]
   \[
   588 = 16.25v^2
   \]
   \[
   v^2 = \frac{588}{16.25}
   \]
   \[
   v \approx 2.73 \, \text{m/s}
   \]

So, the velocity of the block just before it touches the ground is approximately **2.73 m/s**.

---

#### Part (b): Angular velocity of the pulley at that instant

Since \(v = \omega R\):
\[
\omega = \frac{v}{R} = \frac{2.73}{0.1} = 27.3 \, \text{rad/s}
\]

The angular velocity of the pulley at that instant is **27.3 rad/s**.

---

#### Part (c): Tension in the rope

To find the tension, consider the forces acting on the 30 kg block. Let \(T\) be the tension in the rope.

From Newton's second law for the block:
\[
mg - T = ma
\]
where \(a\) is the acceleration of the block. Also, using the rotational dynamics of the pulley, we get:
\[
T \cdot R = I \cdot \alpha
\]
where \(\alpha = \frac{a}{R}\) is the angular acceleration of the pulley. Substituting \(I = \frac{1}{2}MR^2\):
\[
T \cdot R = \frac{1}{2}MR^2 \cdot \frac{a}{R}
\]
\[
T = \frac{1}{2}Ma
\]

Now we can solve for \(a\) and \(T\) by substituting values.

---

#### Part (d): Time for the block to reach the ground

Using \( v = u + at \) and \( v = 2.73 \, \text{m/s}\)

The system shown in the figure is initially at rest. The 30 kg block is 2 m from the ground. The pulley (I = 1/2 MR^2) is a uniform disk with a diameter of 20 cm and a mass of 5 kg. Assume that the rope does not slip on the pulley. Find:

a. The velocity of the 30 kg block just before it hits the ground.
b. The angular velocity of the pulley at that instant.
c. The tension in the rope.
d. The time it takes for the 30 kg block to hit the ground.
e. Solve the problem using dynamics and energy balance.

Answer: 2.73 m/s

Solve a physics problem involving a 30 kg block falling 2 m to the ground connected to a 5 kg pulley. Calculate the block's velocity before hitting the ground, the pulley’s angular velocity, tension in the rope, and time to fall. Apply energy conservation and dynamics principles.

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