A stone is projected with speed of `50 ms^(-1)` at an angle of `60^(@)` with the horizontal. The speed of the stone at highest point of trajectory is

Options

A. `75 ms^(-1)`
B. `25ms^(-1)`
C. `50 ms^(-1)`
D. cannot find

Explanation

To solve the problem, we need to find the speed of the stone at the highest point of its trajectory when it is projected at a speed of 50 m/s at an angle of 60 degrees with the horizontal.

### Step-by-Step Solution:

1. **Identify the Components of Initial Velocity**:
The initial velocity (V₀) of the stone is given as 50 m/s. We can break this velocity into two components: horizontal (Vx) and vertical (Vy).
- Vx = V₀ * cos(θ)
- Vy = V₀ * sin(θ)
where θ is the angle of projection (60 degrees).

2. **Calculate the Horizontal Component (Vx)**:
Using the cosine function:
\[ Vx = 50 \, \text{m/s} \times \cos(60^\circ) \]
Since \(\cos(60^\circ) = \frac{1}{2}\):
\[ Vx = 50 \, \text{m/s} \times \frac{1}{2} = 25 \, \text{m/s} \]

3. **Calculate the Vertical Component (Vy)**:
Using the sine function:
\[ Vy = 50 \, \text{m/s} \times \sin(60^\circ) \]
Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[ Vy = 50 \, \text{m/s} \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \, \text{m/s} \]

4. **Determine the Speed at the Highest Point**:
At the highest point of the trajectory, the vertical component of the velocity (Vy) becomes 0 because the stone momentarily stops rising before it starts to fall. Therefore, the speed at the highest point is only due to the horizontal component (Vx):
\[ \text{Speed at highest point} = Vx = 25 \, \text{m/s} \]

### Final Answer:
The speed of the stone at the highest point of the trajectory is **25 m/s**.