A. Identify the type of motion. B. Calculate the magnitude of the velocity at point B and the height of point X above the ground.

Here's a step-by-step solution to the problem:

Let's define downwards as the positive direction. The acceleration due to gravity is $a = +9.8 \text{ m/s}^2$.

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3.1 Name the type of motion described above.

The object is released from rest and falls under the influence of gravity only, with air friction ignored. This type of motion is called free fall.

Answer for 3.1:
Free fall

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3.2 Calculate the:

3.2.1 Magnitude of the velocity of the object at point B

Step 1: Identify knowns for the motion from B to C.

• Displacement, $\Delta y_{BC} = +30 \text{ m}$ (downwards is positive)
• Time, $t_{BC} = 1.5 \text{ s}$
• Acceleration, $a = +9.8 \text{ m/s}^2$

Step 2: Use the kinematic equation $\Delta y = u t + \frac{1}{2}at^2$ to find the initial velocity at B ($u_B$).

$$30 = u_B (1.5) + \frac{1}{2}(9.8)(1.5)^2$$ $$30 = 1.5 u_B + 4.9(2.25)$$ $$30 = 1.5 u_B + 11.025$$ $$1.5 u_B = 30 - 11.025$$ $$1.5 u_B = 18.975$$ $$u_B = \frac{18.975}{1.5}$$ $$u_B = 12.65 \text{ m/s}$$

The magnitude of the velocity at point B is $\boxed{12.65 \text{ m/s}}$.

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3.2.2 Height of point X above the ground

Step 1: Identify knowns for the motion from X to B.

• Initial velocity at X, $u_X = 0 \text{ m/s}$ (released from rest)
• Final velocity at B, $v_B = 12.65 \text{ m/s}$ (from 3.2.1)
• Acceleration, $a = +9.8 \text{ m/s}^2$

Step 2: Use the kinematic equation $v^2 = u^2 + 2a\Delta y$ to find the displacement from X to B ($\Delta y_{XB}$).

$$(12.65)^2 = (0)^2 + 2(9.8)\Delta y_{XB}$$ $$160.0225 = 19.6 \Delta y_{XB}$$ $$\Delta y_{XB} = \frac{160.0225}{19.6}$$ $$\Delta y_{XB} \approx 8.164 \text{ m}$$

Step 3: Calculate the total height of point X above the ground.
The total height of point X above the ground is the sum of the displacement from X to B and the displacement from B to C.

$$\text{Height of X} = \Delta y_{XB} + \Delta y_{BC}$$ $$\text{Height of X} = 8.164 \text{ m} + 30 \text{ m}$$ $$$$