An arrow is shot into the air at an angle of 59.6° above the horizontal with a speed of 21.5 m/s.

(a) What are the x- and y-components of the velocity of the arrow 3.1 s after it leaves the bowstring?
vx = m/s
vy = m/s
(b) What are the x- and y-components of the displacement of the arrow during the 3.1 s interval?
x = m
y = m

To answer part (a), we need to calculate the x and y components of the velocity of the arrow after 3.1 seconds.

The x-component (horizontal component) of the velocity remains constant throughout the motion because there is no horizontal acceleration. Therefore, the x-component of the velocity at any time is the same as the initial x-component of the velocity.

Given that the arrow was shot at an angle of 59.6° above the horizontal, we can use trigonometry to find the x-component of the initial velocity:

Vx = initial velocity * cos(angle)
= 21.5 m/s * cos(59.6°)

Calculating this, we find:
Vx = 21.5 m/s * 0.527 = 11.32 m/s

So, the x-component of the velocity after 3.1 seconds is also 11.32 m/s.

For the y-component (vertical component) of the velocity, we need to consider the effect of gravity. The arrow will experience a downward acceleration due to gravity. The velocity in the y-direction changes with time due to this acceleration.

The equation for the downward motion (y-direction) of the arrow is given by:
y = initial velocity * time - (1/2) * acceleration * time^2

Given that the initial velocity in the y-direction is:
Vy = initial velocity * sin(angle)
= 21.5 m/s * sin(59.6°)

Calculating this, we find:
Vy = 21.5 m/s * 0.809 = 17.42 m/s

And the vertical acceleration due to gravity is approximately 9.8 m/s^2.

Plugging in the values for time (3.1 seconds), initial velocity (17.42 m/s), and acceleration (-9.8 m/s^2), we can calculate the y-component of the velocity after 3.1 seconds.

Using the equation:
y = Vy * time - (1/2) * acceleration * time^2
= 17.42 m/s * 3.1 s - (1/2) * 9.8 m/s^2 * (3.1 s)^2

Calculating this, we find:
y = 17.42 m/s * 3.1 s - 4.9 m/s^2 * (3.1 s)^2
= 53.962 m - 46.21 m
= 7.752 m/s

So, the y-component of the velocity after 3.1 seconds is 7.752 m/s.

For part (b), we need to calculate the displacement in the x and y directions during the 3.1 second interval.

The displacement in the x-direction is given by:
x = Vx * time
= 11.32 m/s * 3.1 s

Calculating this, we find:
x = 11.32 m/s * 3.1 s
= 35.012 m

So, the displacement in the x-direction during the 3.1 second interval is 35.012 m.

The displacement in the y-direction is given by:
y = Vy * time + (1/2) * acceleration * time^2
= 17.42 m/s * 3.1 s + (1/2) * (-9.8 m/s^2) * (3.1 s)^2

Calculating this, we find:
y = 17.42 m/s * 3.1 s - 4.9 m/s^2 * (3.1 s)^2
= 53.962 m - 46.21 m
= 7.752 m

So, the displacement in the y-direction during the 3.1 second interval is 7.752 m.

So, the final answers are:
(a) vx = 11.32 m/s, vy = 7.752 m/s
(b) x = 35.012 m, y = 7.752 m