Velocity of a stone projected, 2 second bofore it reaches the maximum height makes angle `53^@` with the horizontal then the velocity at highest point will be

Options

A. `20 m//s`
B. `15 m//s`
C. `25 m//s`
D. `80//3 m//s`

Explanation

To solve the problem, we need to find the velocity of a stone at its highest point after it has been projected. We know that the stone's velocity makes an angle of 53° with the horizontal two seconds before it reaches the maximum height.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The stone is projected and reaches its maximum height after a certain time.
- We are given that 2 seconds before reaching the maximum height, the stone's velocity makes an angle of 53° with the horizontal.

2. **Identifying the Components of Velocity**:
- Let \( V \) be the velocity of the stone 2 seconds before it reaches the maximum height.
- The vertical component of the velocity at this point can be expressed as \( V_y \) and the horizontal component as \( V_x \).

3. **Using the Angle to Find Velocity Components**:
- From the angle given, we can use the tangent function:
\[ \tan(53^\circ) = \frac{V_y}{V_x} \]
- This implies:
\[ V_y = V_x \cdot \tan(53^\circ) \]

4. **Finding the Vertical Component of Velocity**:
- The vertical component of the velocity can also be expressed using the equation of motion:
\[ V_y = u_y - g t \]
- Here, \( u_y \) is the initial vertical velocity, \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)), and \( t \) is the time (2 seconds).
- Since at the maximum height \( V_y = 0 \), we can rearrange the equation:
\[ V_y = 0 - g \cdot 2 = -20 \, \text{m/s} \]

5. **Finding the Horizontal Component**:
- Now substituting \( V_y \) into the tangent equation:
\[ -20 = V_x \cdot \tan(53^\circ) \]
- We know \( \tan(53^\circ) \approx 1.6 \):
\[ V_x = \frac{-20}{\tan(53^\circ)} \approx \frac{-20}{1.6} \approx -12.5 \, \text{m/s} \]

6. **Velocity at the Highest Point**:
- At the highest point, the vertical component of the velocity \( V_y \) becomes 0.
- The horizontal component \( V_x \) remains constant throughout the motion (since there is no horizontal acceleration).
- Therefore, the velocity at the highest point is:
\[ V_{max height} = V_x = -12.5 \, \text{m/s} \]
- The negative sign indicates the direction of the velocity vector.

### Final Answer:
The velocity at the highest point will be \( 12.5 \, \text{m/s} \) in the horizontal direction.