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 step by step, we need to analyze the situation involving the car, the parcel, and the man standing on 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**:
- Assume the car is moving in the positive x-direction.
- The man is standing at a point on the side of the road, which we can consider as the origin (0,0).

3. **Determine the velocity of the parcel in the ground frame**:
- When the parcel is thrown at an angle \( \theta \) with respect to the direction of the car, its velocity in the ground frame can be decomposed into two components:
- Horizontal component: \( V_{px} = V_p \cos(\theta) \)
- Vertical component: \( V_{py} = V_p \sin(\theta) \)

- The total horizontal velocity of the parcel in the ground frame is:
\[ V_{px} = V_c + V_p \cos(\theta) \]
- The vertical component remains:
\[ V_{py} = V_p \sin(\theta) \]

4. **Condition for the parcel to reach the man**:
- For the parcel to reach the man standing at the origin, it must have no horizontal displacement relative to the man when it reaches the ground. This means that the horizontal component of the parcel's velocity must equal the velocity of the car:
\[ V_{px} = 0 \]
- Thus, we have:
\[ V_c + V_p \cos(\theta) = 0 \]

5. **Substituting the known values**:
- Plugging in the values:
\[ 10 + 10\sqrt{2} \cos(\theta) = 0 \]
- Rearranging gives:
\[ \cos(\theta) = -\frac{10}{10\sqrt{2}} = -\frac{1}{\sqrt{2}} \]

6. **Finding the angle**:
- The angle \( \theta \) for which \( \cos(\theta) = -\frac{1}{\sqrt{2}} \) corresponds to:
\[ \theta = 180^\circ - 45^\circ = 135^\circ \]

7. **Final answer**:
- Therefore, the angle at which the parcel must be thrown with respect to the direction of motion of the car is:
\[ \theta = 135^\circ \]