An arrow is shot into the air at an angle of 59.4° above the horizontal with a speed of 20.0 m/s. What is the x-component of the velocity of the arrow 3.30 s after it leaves the bowstring?

What is the y-component of the velocity of the arrow 3.30 s after it leaves the bowstring?
What is the x-component of the displacement of the arrow during the 3.30-s interval?
What is the y-component of the displacement of the arrow during the 3.30-s interval?

Vo = 20m/s[59.4o]

Xo = 20*cos59.4 = 10.18 m/s.
Yo = 20*sin59.4 = 17.21 m/s.

X = Xo = 10.18 m/s. Does not change.

Y = Yo+g*t = 17.21 - 9.8*3.30=-15.13 m/s
Downward.

Dx = Xo*T = 10.18 * 3.30 = 33.6 m.

h = (Y^2-Yo^2)/2g
h = (-15.13)^2-(17.21^2)/-19.6 =
( 228.92-296.18)/-19.6 = 3.43 m. = Y.

To solve these questions, we can use the equations of motion for projectile motion:

1. The x-component of the velocity of the arrow after 3.30 s can be found using the equation:
Vx = V₀ * cosθ
where V₀ is the initial velocity and θ is the angle of projection.
Plugging in the values, we have:
Vx = 20.0 m/s * cos(59.4°)
Vx ≈ 20.0 m/s * 0.557
Vx ≈ 11.14 m/s

Therefore, the x-component of the velocity of the arrow after 3.30 s is approximately 11.14 m/s.

2. The y-component of the velocity of the arrow after 3.30 s can be found using the equation:
Vy = V₀ * sinθ - g * t
where g is the acceleration due to gravity (9.8 m/s²) and t is the time.
Plugging in the values, we have:
Vy = 20.0 m/s * sin(59.4°) - (9.8 m/s²) * 3.30 s
Vy ≈ 20.0 m/s * 0.841 - 32.34 m/s
Vy ≈ 16.82 m/s - 32.34 m/s
Vy ≈ -15.52 m/s

Therefore, the y-component of the velocity of the arrow after 3.30 s is approximately -15.52 m/s. Note that the negative sign indicates that the arrow is moving downwards.

3. The x-component of the displacement of the arrow during the 3.30-s interval can be found using the equation:
Dx = Vx * t
where Vx is the x-component of the velocity and t is the time.
Plugging in the values, we have:
Dx = (11.14 m/s) * (3.30 s)
Dx ≈ 36.682 m

Therefore, the x-component of the displacement of the arrow during the 3.30-s interval is approximately 36.682 m.

4. The y-component of the displacement of the arrow during the 3.30-s interval can be found using the equation:
Dy = Vy * t + (1/2) * g * t²
where Vy is the y-component of the velocity, g is the acceleration due to gravity, and t is the time.
Plugging in the values, we have:
Dy = (-15.52 m/s) * (3.30 s) + (1/2) * (9.8 m/s²) * (3.30 s)²
Dy ≈ -51.216 m + 53.914 m
Dy ≈ 2.698 m

Therefore, the y-component of the displacement of the arrow during the 3.30-s interval is approximately 2.698 m. Note that the positive value indicates that the arrow has moved upwards.

To solve these problems, we need to analyze the motion of the arrow using the principles of projectile motion.

1. X-component of velocity:
The x-component of the velocity represents the horizontal velocity of the arrow. Since there are no horizontal forces acting on the arrow (assuming no air resistance), the horizontal velocity remains constant throughout the motion. In this case, the given horizontal velocity is 20.0 m/s.

2. Y-component of velocity:
The y-component of the velocity represents the vertical velocity of the arrow. The vertical velocity changes due to the acceleration due to gravity. The initial vertical velocity, vy0, can be calculated using the formula vy0 = v * sin(θ), where v is the initial speed (20.0 m/s) and θ is the angle (59.4°).
Once we find the initial vertical velocity, we can calculate the y-component of velocity at any given time using the equation vy = vy0 - g * t, where g is the acceleration due to gravity (9.8 m/s²) and t is the time (3.30 s).

3. X-component of displacement:
The x-component of displacement represents the horizontal distance covered by the arrow during the given time interval. Since the horizontal velocity is constant, the x-component of displacement can be calculated as dx = vx * t, where vx is the x-component of velocity (20.0 m/s) and t is the time (3.30 s).

4. Y-component of displacement:
The y-component of displacement represents the vertical distance covered by the arrow during the given time interval. The vertical distance can be calculated using the formula dy = vy0 * t + (1/2) * g * t², where vy0 is the initial vertical velocity obtained in step 2 and g is the acceleration due to gravity (9.8 m/s²).

By following these steps, you should be able to calculate the x-component of the velocity, y-component of the velocity, x-component of the displacement, and y-component of the displacement of the arrow during the given time interval.