An arrow is shot into the air at an angle of 60.3° above the horizontal with a speed of 21.2 m/s. What are the x- and y-components of the velocity of the arrow 3.1 s after it leaves the bowstring? What are the x- and y-components of the displacement of the arrow during the 3.1 s interval?
To find the x- and y-components of the velocity of the arrow, we can use the following equations:
Vx = V * cos(θ)
Vy = V * sin(θ)
Where:
- Vx is the x-component of the velocity
- Vy is the y-component of the velocity
- V is the magnitude of the velocity (21.2 m/s in this case)
- θ is the angle above the horizontal (60.3° in this case)
First, let's calculate the x-component of the velocity:
Vx = 21.2 m/s * cos(60.3°)
To get the value of cos(60.3°), we can use a scientific calculator or look it up in a trigonometric table. The value of cos(60.3°) is approximately 0.5.
Vx ≈ 21.2 m/s * 0.5
Vx ≈ 10.6 m/s
Now, let's calculate the y-component of the velocity:
Vy = 21.2 m/s * sin(60.3°)
Similarly, we need the value of sin(60.3°) which is approximately 0.866.
Vy ≈ 21.2 m/s * 0.866
Vy ≈ 18.36 m/s
So, the x-component of the velocity is approximately 10.6 m/s, and the y-component of the velocity is approximately 18.36 m/s.
To find the x- and y-components of the displacement of the arrow during the 3.1 s interval, we can use the following equations:
Dx = Vx * t
Dy = Vy * t + (1/2) * g * t^2
Where:
- Dx is the x-component of the displacement
- Dy is the y-component of the displacement
- t is the time interval (3.1 s in this case)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
First, let's calculate the x-component of the displacement:
Dx = 10.6 m/s * 3.1 s
Dx ≈ 32.86 m
Now, let's calculate the y-component of the displacement:
Dy = 18.36 m/s * 3.1 s + (1/2) * 9.8 m/s^2 * (3.1 s)^2
Dy ≈ 178.34 m
So, the x-component of the displacement is approximately 32.86 m, and the y-component of the displacement is approximately 178.34 m.