c): Range d): velocity and direction of the motion at a height of 130 meters

Step 1: Identify the given values and formulas.
Given:
Initial velocity, $u = 196 \text{ m/s}$
Angle of projection, $\theta = 30^\circ$
Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

c) Calculate the Range.
The formula for the horizontal range ($R$) is:

$$R = \frac{u^2 \sin(2\theta)}{g}$$

Substitute the given values:

$$R = \frac{(196 \text{ m/s})^2 \sin(2 \times 30^\circ)}{9.8 \text{ m/s}^2}$$ $$R = \frac{(196 \text{ m/s})^2 \sin(60^\circ)}{9.8 \text{ m/s}^2}$$ $$R = \frac{38416 \text{ m}^2/\text{s}^2 \times \frac{\sqrt{3}}{2}}{9.8 \text{ m/s}^2}$$ $$R = \frac{38416 \times 0.866025}{9.8} \text{ m}$$ $$R = \frac{33269.96}{9.8} \text{ m}$$ $$R \approx 3394.89 \text{ m}$$

The range is $\boxed{\text{3395 m}}$.

d) Calculate the velocity and direction of the motion at a height of 130 meters.
First, find the horizontal and vertical components of the initial velocity:
Horizontal component: $u_x = u \cos \theta = 196 \cos 30^\circ = 196 \times \frac{\sqrt{3}}{2} = 98\sqrt{3} \text{ m/s}$
Vertical component: $u_y = u \sin \theta = 196 \sin 30^\circ = 196 \times 0.5 = 98 \text{ m/s}$

The horizontal velocity ($v_x$) remains constant throughout the flight:

$$v_x = u_x = 98\sqrt{3} \text{ m/s} \approx 169.71 \text{ m/s}$$

To find the vertical velocity ($v_y$) at a height $h = 130 \text{ m}$, use the kinematic equation:

$$v_y^2 = u_y^2 - 2gh$$ $$v_y^2 = (98 \text{ m/s})^2 - 2 \times 9.8 \text{ m/s}^2 \times 130 \text{ m}$$ $$v_y^2 = 9604 \text{ m}^2/\text{s}^2 - 2548 \text{ m}^2/\text{s}^2$$ $$v_y^2 = 7056 \text{ m}^2/\text{s}^2$$ $$v_y = \pm \sqrt{7056} \text{ m/s}$$ $$v_y = \pm 84 \text{ m/s}$$

The positive value corresponds to the particle moving upwards, and the negative value corresponds to it moving downwards.

The magnitude of the velocity ($v$) at this height is:

$$v = \sqrt{v_x^2 + v_y^2}$$ $$v = \sqrt{(98\sqrt{3} \text{ m/s})^2 + (\pm 84 \text{ m/s})^2}$$ $$v = \sqrt{(9604 \times 3) + 7056} \text{ m/s}$$ $$v = \sqrt{28812 + 7056} \text{ m/s}$$ $$v = \sqrt{35868} \text{ m/s}$$ $$v \approx 189.39 \text{ m/s}$$

The magnitude of the velocity is $\boxed{\text{189.39 m/s}}$.

The direction of motion ($\alpha$) is given by:

$$\tan \alpha = \frac{v_y}{v_x}$$

If the particle is moving upwards ($v_y = 84 \text{ m/s}$):

$$\tan \alpha = \frac{84 \text{ m/s}}{98\sqrt{3} \text{ m/s}} \approx \frac{84}{169.71} \approx 0.4950$$ $$\alpha = \arctan(0.4950) \approx 26.35^\circ$$

The direction is $\boxed{\text{26.35}^\circ \text{ above the horizontal}}$.

If the particle is moving downwards ($v_y = -84 \text{ m/s}$):

$$\tan \alpha = \frac{-84 \text{ m/s}}{98\sqrt{3} \text{ m/s}} \approx \frac{-84}{169.71} \approx -0.4950$$ $$\alpha = \arctan(-0.4950) \approx -26.35^\circ$$

The direction is $\boxed{\text{26.35}^\circ \text{ below the horizontal}}$.