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 of finding the speed of a stone at the highest point of its trajectory when projected with a speed of 50 m/s at an angle of 60 degrees with the horizontal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Velocity Components**:
The initial velocity \( u \) can be broken down into horizontal and vertical components using trigonometric functions:
- Horizontal component \( V_x = u \cos \theta \)
- Vertical component \( V_y = u \sin \theta \)

Given:
- \( u = 50 \, \text{m/s} \)
- \( \theta = 60^\circ \)

Calculate:
\[ V_x = 50 \cos(60^\circ) = 50 \times \frac{1}{2} = 25 \, \text{m/s} \]
\[ V_y = 50 \sin(60^\circ) = 50 \times \frac{\sqrt{3}}{2} \approx 43.3 \, \text{m/s} \]

2. **Understand the Motion at the Highest Point**:
At the highest point of the trajectory, the vertical component of the velocity \( V_y \) becomes zero due to the effect of gravity. However, the horizontal component \( V_x \) remains constant throughout the motion because there are no horizontal forces acting on the stone.

3. **Determine the Speed at the Highest Point**:
At the highest point, the speed of the stone will only be due to the horizontal component:
\[ V_{\text{highest point}} = V_x = 25 \, \text{m/s} \]

4. **Conclusion**:
Therefore, the speed of the stone at the highest point of its trajectory is \( 25 \, \text{m/s} \).

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

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