A car is travelling at a velocity `"10 km h"^(-1)` on a straight road. The driver of the car throws a parcel with a velocity `10sqrt2km h^(-1)` with respect to the car, when the car is passing by a man standing on the side of the road. If the parcel is to reach the man, the direction of throw makes the following angle with the direction of motion of the car

Options

A. `135^(@)`
B. `45^(@)`
C. `tan^(-1)sqrt2`
D. `tan^(-1)((1)/(sqrt2))`

Answer

To solve the problem, we need to analyze the motion of the parcel thrown from the car and determine the angle at which the parcel must be thrown so that it reaches the man standing by the side of the road.

### Step-by-Step Solution:

1. **Identify the velocities:**
- The velocity of the car, \( V_c = 10 \, \text{km/h} \).
- The velocity of the parcel with respect to the car, \( V_p = 10\sqrt{2} \, \text{km/h} \).

2. **Set up the coordinate system:**
- Let the direction of the car's motion be along the positive y-axis.
- The parcel is thrown at an angle \( \alpha \) with respect to the direction of the car's motion.

3. **Resolve the velocity of the parcel:**
- The velocity of the parcel can be resolved into two components:
- Along the y-axis (direction of the car): \( V_{py} = V_p \sin(\alpha) \)
- Along the x-axis (perpendicular to the car's motion): \( V_{px} = V_p \cos(\alpha) \)

4. **Determine the condition for the parcel to reach the man:**
- For the parcel to reach the man, the resultant velocity of the parcel in the y-direction must be zero (since the man is stationary):
\[ V_{py} - V_c = 0 \implies V_{py} = V_c \]
- Therefore, we have:
\[ V_p \sin(\alpha) = V_c \]

5. **Substituting the known values:**
- Substitute \( V_p = 10\sqrt{2} \, \text{km/h} \) and \( V_c = 10 \, \text{km/h} \):
\[ 10\sqrt{2} \sin(\alpha) = 10 \]
- Dividing both sides by 10:
\[ \sqrt{2} \sin(\alpha) = 1 \implies \sin(\alpha) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

6. **Finding the angle \( \alpha \):**
- The angle \( \alpha \) that satisfies this equation is:
\[ \alpha = 45^\circ \]

7. **Determine the angle \( \theta \) with respect to the direction of motion of the car:**
- The angle \( \theta \) that the line of throw makes with the direction of motion of the car is:
\[ \theta = 90^\circ + \alpha = 90^\circ + 45^\circ = 135^\circ \]

### Final Answer:
The angle at which the parcel must be thrown with respect to the direction of motion of the car is \( 135^\circ \).