A motor boat is to reach at point `30^(@)` upsteam (w.rt. Normal) on other side of a river flowing with velocity `5m//s`. Velocity of motorboat w.r.t. water is `5sqrt(3) m//s`. The driver should steer the baot at an angular

Options

A. `120^(@)`w.r.t. stream direction
B. `30^(@)` w.r.t. normal to the bank
C. `30^(@)` w.r.t. the line of destination form starting point.
D. none of these

Explanation

To solve the problem of the motorboat reaching a point 30 degrees upstream while navigating a river flowing at 5 m/s, we can break down the problem into steps.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Velocity of the river, \( V_r = 5 \, \text{m/s} \)
- Velocity of the motorboat w.r.t water, \( V_b = 5\sqrt{3} \, \text{m/s} \)
- Angle to reach upstream, \( \theta = 30^\circ \)

2. **Set Up the Coordinate System:**
- Let the positive x-direction be downstream (in the direction of the river flow).
- Let the positive y-direction be perpendicular to the river flow (across the river).

3. **Break Down the Velocity of the Motorboat:**
- The velocity of the motorboat can be broken down into two components:
- \( V_{bx} = V_b \cdot \cos(\theta) \) (upstream component)
- \( V_{by} = V_b \cdot \sin(\theta) \) (across the river component)

4. **Account for the River's Velocity:**
- The effective velocity of the boat upstream must counteract the river's flow. Thus, the net upstream velocity \( V_{net} \) can be expressed as:
\[ V_{net} = V_{bx} - V_r \]
- For the boat to reach the target point at a 30-degree angle upstream, we need the ratio of the vertical and horizontal components of the boat's velocity to satisfy:
\[ \tan(30^\circ) = \frac{V_{by}}{V_{net}} \]

5. **Substituting Values:**
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).
- Therefore, we can write:
\[ \frac{1}{\sqrt{3}} = \frac{V_{by}}{V_{bx} - V_r} \]

6. **Expressing Components:**
- Substitute the expressions for \( V_{by} \) and \( V_{bx} \):
\[ V_{by} = V_b \cdot \sin(\theta) = 5\sqrt{3} \cdot \sin(\theta) \]
\[ V_{bx} = V_b \cdot \cos(\theta) = 5\sqrt{3} \cdot \cos(\theta) \]
- Thus, the equation becomes:
\[ \frac{1}{\sqrt{3}} = \frac{5\sqrt{3} \cdot \sin(\theta)}{5\sqrt{3} \cdot \cos(\theta) - 5} \]

7. **Simplifying the Equation:**
- Cancel \( 5\sqrt{3} \) from both sides:
\[ \frac{1}{\sqrt{3}} = \frac{\sin(\theta)}{\cos(\theta) - 1} \]
- Cross-multiplying gives:
\[ \cos(\theta) - 1 = \sqrt{3} \sin(\theta) \]

8. **Using Trigonometric Identities:**
- Rearranging gives:
\[ \cos(\theta) - \sqrt{3} \sin(\theta) = 1 \]
- This equation can be solved for \( \theta \).

9. **Finding the Angle \( \theta \):**
- Solving the equation will yield \( \theta = 30^\circ \).

### Final Answer:
The driver should steer the boat at an angle of \( 30^\circ \) upstream.