Two particles are thrown up simultaneously with a velocity of 30m/s,one thrown vertically and another at 45° with respect to the horizon.What is the distance between them at t=1.5s

To find the distance between the two particles after 1.5 seconds, we need to analyze their motions separately. One particle is thrown vertically, while the other is thrown at a 45° angle. We'll break down their movements into horizontal and vertical components and then calculate their positions at the given time.

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 after 1.5 seconds:

• Equation: s = ut + (1/2)at²

Here, s is the displacement, u is the initial velocity (30 m/s), a is the acceleration due to gravity (approximately -9.81 m/s², acting downwards), and t is the time (1.5 s).

Substituting the values:

• s = 30 m/s * 1.5 s + (1/2)(-9.81 m/s²)(1.5 s)²
• s = 45 m - 11.03 m
• s ≈ 33.97 m

Motion of the Second Particle

The second particle is thrown at a 45° angle. We need to find its horizontal and vertical components of the initial velocity:

• Horizontal Component: u_x = u * cos(θ) = 30 m/s * cos(45°) = 30 m/s * (√2/2) ≈ 21.21 m/s
• Vertical Component: u_y = u * sin(θ) = 30 m/s * sin(45°) = 30 m/s * (√2/2) ≈ 21.21 m/s

Calculating the Position of the Second Particle

Now, we can find the position of the second particle after 1.5 seconds using the same kinematic equation for both components:

Horizontal Position

• x = u_x * t = 21.21 m/s * 1.5 s ≈ 31.82 m

Vertical Position

• y = u_y * t + (1/2)(-9.81 m/s²)(1.5 s)²
• y = 21.21 m/s * 1.5 s - 11.03 m ≈ 21.21 m/s * 1.5 s - 11.03 m ≈ 21.21 m - 11.03 m ≈ 10.18 m

Finding the Distance Between the Two Particles

Now that we have the positions of both particles, we can determine the distance between them. The first particle is at (0, 33.97 m) and the second particle is at (31.82 m, 10.18 m). We can use the distance formula:

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

Substituting the coordinates:

• d = √((31.82 m - 0)² + (10.18 m - 33.97 m)²)
• d = √((31.82)² + (-23.79)²)
• d = √(1011.0724 + 566.0641) ≈ √1577.1365 ≈ 39.7 m

Thus, the distance between the two particles at t = 1.5 seconds is approximately 39.7 meters.