A stone is catapulted at time t = 0, with an initial velocity of magnitude 21.5 m/s and at an angle of 35.3° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.06 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.73 s.

To find the magnitudes of the horizontal and vertical components of displacement at different times, we need to use the equations of motion for projectile motion.

Let's break down the problem step by step:

Step 1: Split the initial velocity into horizontal and vertical components.
The initial velocity has a magnitude of 21.5 m/s and is at an angle of 35.3° above the horizontal. To find the horizontal and vertical components, we can use trigonometry.

Horizontal Component: Vx = V * cos(θ)
Vx = 21.5 m/s * cos(35.3°)
Vx ≈ 17.571 m/s

Vertical Component: Vy = V * sin(θ)
Vy = 21.5 m/s * sin(35.3°)
Vy ≈ 12.007 m/s

Step 2: Find the horizontal displacement at different times.
The horizontal displacement (Δx) is given by the equation:

Δx = Vx * t

(a) At t = 1.06 s:
Δx = 17.571 m/s * 1.06 s
Δx ≈ 18.623 m

(c) At t = 1.73 s:
Δx = 17.571 m/s * 1.73 s
Δx ≈ 30.407 m

Step 3: Find the vertical displacement at different times.
The vertical displacement (Δy) is given by the equation:

Δy = Vy * t + (1/2) * g * t^2

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

(b) At t = 1.06 s:
Δy = 12.007 m/s * 1.06 s + (1/2) * 9.8 m/s^2 * (1.06 s)^2
Δy ≈ 9.754 m

(d) At t = 1.73 s:
Δy = 12.007 m/s * 1.73 s + (1/2) * 9.8 m/s^2 * (1.73 s)^2
Δy ≈ 16.366 m

So, the magnitudes of the (a) horizontal and (b) vertical components of the displacement at t = 1.06 s are approximately 18.623 m and 9.754 m, respectively.

Similarly, the magnitudes of the (c) horizontal and (d) vertical components of the displacement at t = 1.73 s are approximately 30.407 m and 16.366 m, respectively.