https://file.aiquickdraw.com/mathbot/file/172638993242574vvsg0h.jpg

The question in the image asks about a particle projected at an angle of 60° with the horizontal, moving along a parabolic trajectory, and reaching the top point of the trajectory.

To solve this, let’s break down the problem:

### Key Information:
- Angle of projection, \(\theta = 60^\circ\)
- The particle has a constant speed equal to the initial speed.
- Gravity acceleration, \(g = 10 \, \text{m/s}^2\).
- The question asks for the acceleration at the top of the trajectory.

### Important Concept:
At the top of a projectile's trajectory, the vertical component of velocity becomes zero, and only the horizontal component remains. In this case, since there’s no vertical motion, the only acceleration acting on the particle is due to gravity, which points downward.

### Acceleration at the Top:
- Since the motion is only horizontal at the top, the only force affecting the particle is gravity.
- Hence, the magnitude of acceleration will be the gravitational acceleration itself, which is:
\[
a = g = 10 \, \text{m/s}^2
\]

### Final Answer:
The magnitude of the acceleration at the top of the trajectory is \( \boxed{10} \, \text{m/s}^2 \).

Would you like any more detailed steps or clarifications?

### Related Questions:
1. How would the acceleration change if the projection angle were different?
2. What is the velocity of the particle at the top of its trajectory?
3. How do you calculate the time to reach the top of the trajectory?
4. What are the horizontal and vertical components of the initial velocity?
5. How does air resistance affect the acceleration at the top?

### Tip:
In projectile motion, remember that the horizontal acceleration is always zero (neglecting air resistance), and vertical acceleration is always \(g\), directed downward.

A ball is projected at angle 60° with horizontal. A particle is made to move along the trajectory of ball with constant speed equal to initial speed of ball. The magnitude of acceleration (in m/s^2) of the particle at the top of trajectory is ——— (take g = 10 m/s^2).

Learn how to calculate the acceleration of a particle at the top of its trajectory when projected at an angle of 60° with the horizontal. This solution covers key kinematic concepts and the effect of gravity. Suitable for students in Grades 9-12.

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