two particles are thrownup simultaneously with a velocity of 30 m/s and one is thrown vertically upward and another at 45 degree with respect to the horizon.Find out distance between them at time=1.5 s

To find the distance between the two particles after 1.5 seconds, we need to analyze their motions separately. One particle is thrown vertically upward, while the other is thrown at a 45-degree angle. Let's break down the problem step by step.

Vertical Motion of the First Particle

The first particle is thrown straight up with an initial velocity of 30 m/s. We can use the following kinematic equation to determine its position at any time:

• Equation: y = v₀t - (1/2)gt²

Here, y is the vertical position, v₀ is the initial velocity (30 m/s), g is the acceleration due to gravity (approximately 9.81 m/s²), and t is the time in seconds.

Substituting the values for t = 1.5 s:

• y = 30(1.5) - (1/2)(9.81)(1.5)²
• y = 45 - (1/2)(9.81)(2.25)
• y = 45 - 11.03625
• y ≈ 33.96 m

Motion of the Second Particle

The second particle is thrown at a 45-degree angle. We need to find its horizontal and vertical components of motion. The initial velocity can be broken down as follows:

• Horizontal Component: v₀x = v₀ * cos(θ) = 30 * cos(45°) = 30 * (√2/2) ≈ 21.21 m/s
• Vertical Component: v₀y = v₀ * sin(θ) = 30 * sin(45°) = 30 * (√2/2) ≈ 21.21 m/s

Now, we can calculate the position of the second particle after 1.5 seconds using the same kinematic equations:

Vertical Position of the Second Particle

• y = v₀yt - (1/2)gt²
• y = 21.21(1.5) - (1/2)(9.81)(1.5)²
• y = 31.815 - 11.03625
• y ≈ 20.78 m

Horizontal Position of the Second Particle

• x = v₀xt
• x = 21.21(1.5)
• x ≈ 31.82 m

Calculating the Distance Between the Two Particles

Now we have the positions of both particles after 1.5 seconds:

• First Particle: (0, 33.96 m)
• Second Particle: (31.82 m, 20.78 m)

To find the distance between the two particles, we can use the distance formula:

• Distance: d = √((x₂ - x₁)² + (y₂ - y₁)²)

Substituting the coordinates:

• d = √((31.82 - 0)² + (20.78 - 33.96)²)
• d = √((31.82)² + (-13.18)²)
• d = √(1011.0724 + 174.5924)
• d = √(1185.6648)
• d ≈ 34.4 m

Thus, the distance between the two particles after 1.5 seconds is approximately 34.4 meters. This approach illustrates how to analyze projectile motion by breaking it down into its components and applying kinematic equations effectively.