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

To solve the problem of finding the distance between two particles thrown simultaneously, we need to analyze their motions separately and then calculate the distance between them at a specific time, which is 1.5 seconds in this case. Let's break it down step by step.

Analyzing the Vertical Motion

The first particle is thrown straight up with an initial velocity of 30 m/s. To determine its position after 1.5 seconds, we can use the following kinematic equation:

y = v₀t - (1/2)gt²

• y = vertical displacement
• v₀ = initial velocity (30 m/s)
• g = acceleration due to gravity (approximately 9.81 m/s²)
• t = time (1.5 s)

Plugging in the values:

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

Examining the Projectile Motion

The second particle is thrown at a 45-degree angle with the same initial velocity of 30 m/s. We need to find its horizontal and vertical components of the velocity:

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

Now, we can calculate the horizontal and vertical positions after 1.5 seconds:

Horizontal Position

The horizontal position (x) can be calculated using:

x = v₀x * t

x = 21.21 * 1.5 ≈ 31.82 m

Vertical Position

The vertical position (y) for the second particle is calculated similarly:

y = v₀yt - (1/2)gt²

y = 21.21(1.5) - (1/2)(9.81)(1.5)²

y = 31.82 - 11.03625

y ≈ 20.78 m

Finding the Distance Between the Two Particles

Now that we have the positions of both particles at t = 1.5 seconds, we can find the distance between them. The first particle is at (0, 33.96) and the second particle is at (31.82, 20.78). We can use the distance formula:

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

• (x₁, y₁) = (0, 33.96)
• (x₂, y₂) = (31.82, 20.78)

Substituting the values:

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

Summary of Results

At t = 1.5 seconds, the distance between the two particles is approximately 34.4 meters. This calculation illustrates how to analyze both vertical and horizontal motions separately and then combine them to find the distance between two moving objects.