To solve the problem of the projectile fired from the origin, we need to break down the motion into its horizontal and vertical components. Given the initial speed of 50 m/s and the launch angle of 53°, we can use some basic physics equations to find the answers to your questions about the horizontal separation between points A and B, and the distance from the origin at t = 2 seconds.

Projectile Motion Components

First, let's determine the horizontal and vertical components of the initial velocity:

• Horizontal Component (Vx): Vx = V * cos(θ) = 50 * cos(53°)
• Vertical Component (Vy): Vy = V * sin(θ) = 50 * sin(53°)

Calculating these values:

• Vx = 50 * 0.6018 ≈ 30.09 m/s
• Vy = 50 * 0.7986 ≈ 39.93 m/s

Finding Time to Reach Height of 75 m

Next, we need to find the time it takes for the projectile to reach a height of 75 m. We can use the vertical motion equation:

y = Vy * t - (1/2) * g * t²

Where g is the acceleration due to gravity (approximately 9.81 m/s²). Setting y = 75 m, we have:

75 = 39.93 * t - (1/2) * 9.81 * t²

This simplifies to:

0 = -4.905t² + 39.93t - 75

Using the quadratic formula, t = [−b ± √(b² - 4ac)] / 2a, where a = -4.905, b = 39.93, and c = -75:

Calculating the discriminant:

b² - 4ac = (39.93)² - 4 * (-4.905) * (-75) ≈ 1595.6049 - 1477.5 = 118.1049

Now, substituting into the quadratic formula:

t = [−39.93 ± √(118.1049)] / (2 * -4.905)

Calculating the two possible times:

t₁ ≈ 8.96 seconds and t₂ ≈ 1.67 seconds (the time to reach the height on the way up and down).

Horizontal Separation Between Points A and B

Now, we can find the horizontal distance traveled during these times:

Horizontal Distance (d) = Vx * t

For t₁ (upward path):

d₁ = 30.09 * 1.67 ≈ 50.23 m

For t₂ (downward path):

d₂ = 30.09 * 8.96 ≈ 269.45 m

The horizontal separation between points A and B is:

Horizontal Separation = d₂ - d₁ ≈ 269.45 - 50.23 ≈ 219.22 m

However, since we are looking for the separation between the two points at the same height, we only consider the time it takes to reach 75 m on the way up and down, which gives us a total horizontal distance of:

Horizontal Separation = 30.09 * (t₁ + t₂) ≈ 30.09 * (1.67 + 1.67) ≈ 30.09 * 3.34 ≈ 100.4 m.

Since this does not match any of the options provided, we conclude that the answer is (D) None.

Distance from Origin at t = 2 seconds

Now, let’s find the distance of the particle from the origin at t = 2 seconds. We can use the following formulas:

Horizontal Distance (x) = Vx * t

Vertical Distance (y) = Vy * t - (1/2) * g * t²

Calculating x:

x = 30.09 * 2 ≈ 60.18 m

Calculating y:

y = 39.93 * 2 - (1/2) * 9.81 * (2)² ≈ 79.86 - 19.62 ≈ 60.24 m

Now, we can find the resultant distance from the origin using the Pythagorean theorem:

Distance = √(x² + y²)

Distance = √((60.18)² + (60.24)²) ≈ √(3621.6324 + 3628.0576) ≈ √7249.69 ≈ 85.1 m.

Since this also does not match the provided options, we conclude that the answer is (D) None for the distance from the origin at t = 2 seconds as well.