A stone is projectef from level ground such that its horizontal and vertical components of initial velocity are `u_x=10(m)/(s)` and `u_y=20(m)/(s)` respectively. Then the angle between velocity vector of stone one second before and one second after it attains maximum height is:

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `90^@`

Explanation

To find the angle between the velocity vector of the stone one second before and one second after it attains maximum height, we can follow these steps:

### Step 1: Understand the motion of the stone
The stone is projected with horizontal and vertical components of initial velocity:
- \( u_x = 10 \, \text{m/s} \) (horizontal component)
- \( u_y = 20 \, \text{m/s} \) (vertical component)

### Step 2: Determine the time to reach maximum height
At maximum height, the vertical component of the velocity becomes zero. The time to reach maximum height can be calculated using the formula:
\[ t_{max} = \frac{u_y}{g} \]
where \( g \approx 10 \, \text{m/s}^2 \). Thus,
\[ t_{max} = \frac{20}{10} = 2 \, \text{s} \]

### Step 3: Find the velocity vectors one second before and after maximum height
1. **One second before maximum height (at \( t = 1 \, \text{s} \))**:
- The vertical component of the velocity at this time can be calculated using:
\[ v_y = u_y - g \cdot t = 20 - 10 \cdot 1 = 10 \, \text{m/s} \]
- The horizontal component remains constant:
\[ v_x = u_x = 10 \, \text{m/s} \]
- Therefore, the velocity vector one second before maximum height is:
\[ \vec{v}_{before} = (10, 10) \, \text{m/s} \]

2. **One second after maximum height (at \( t = 3 \, \text{s} \))**:
- The vertical component of the velocity at this time can be calculated using:
\[ v_y = u_y - g \cdot t = 20 - 10 \cdot 3 = -10 \, \text{m/s} \]
- The horizontal component remains constant:
\[ v_x = u_x = 10 \, \text{m/s} \]
- Therefore, the velocity vector one second after maximum height is:
\[ \vec{v}_{after} = (10, -10) \, \text{m/s} \]

### Step 4: Calculate the angle between the two velocity vectors
To find the angle \( \theta \) between the two vectors \( \vec{v}_{before} \) and \( \vec{v}_{after} \), we can use the dot product formula:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta) \]
Where:
- \( \vec{A} = (10, 10) \)
- \( \vec{B} = (10, -10) \)

Calculating the dot product:
\[ \vec{A} \cdot \vec{B} = 10 \cdot 10 + 10 \cdot (-10) = 100 - 100 = 0 \]

Since the dot product is zero, the angle between the two vectors is:
\[ \theta = 90^\circ \]

### Final Answer
The angle between the velocity vector of the stone one second before and one second after it attains maximum height is \( 90^\circ \).

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