a particle is projected at an angle of 60 above the horizontal with a speed of 10 m/s.after some time the direction of its velocity makes an angle of 30 above the horizontal.the speed of the particle at this instant is

To determine the speed of the particle at the moment its velocity makes an angle of 30 degrees above the horizontal, we can break down the problem using the principles of projectile motion. Let's analyze the situation step by step.

Initial Conditions

The particle is projected at an angle of 60 degrees with an initial speed of 10 m/s. We can decompose this initial velocity into its horizontal and vertical components:

• Initial Horizontal Velocity (Vx0): Vx0 = V * cos(θ) = 10 m/s * cos(60°) = 10 m/s * 0.5 = 5 m/s
• Initial Vertical Velocity (Vy0): Vy0 = V * sin(θ) = 10 m/s * sin(60°) = 10 m/s * (√3/2) ≈ 8.66 m/s

Velocity Components Over Time

As the particle moves, its horizontal velocity remains constant (assuming no air resistance), while the vertical velocity changes due to gravity. The vertical velocity at any time \( t \) can be expressed as:

Vy(t) = Vy0 - g * t

where \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).

Finding the Speed at 30 Degrees

At the moment when the velocity makes an angle of 30 degrees above the horizontal, we can relate the vertical and horizontal components of the velocity:

tan(30°) = Vy / Vx

From trigonometry, we know that:

• tan(30°) = 1/√3

Thus, we can express the relationship between the vertical and horizontal components at this instant:

Vy = (1/√3) * Vx

Expressing Velocities

We already know that the horizontal velocity \( Vx \) remains constant at 5 m/s. Therefore:

Vy = (1/√3) * 5 m/s ≈ 2.89 m/s

Calculating the Resultant Speed

Now, we can find the resultant speed of the particle using the Pythagorean theorem:

Speed = √(Vx² + Vy²)

Substituting the values we have:

Speed = √(5² + (2.89)²) = √(25 + 8.35) = √33.35 ≈ 5.77 m/s

Final Answer

Therefore, the speed of the particle at the instant when its velocity makes an angle of 30 degrees above the horizontal is approximately 5.77 m/s.