An arrow is shot into the air at an angle of 60.3° above the horizontal with a speed of 20.1 m/s. What are the x- and y-components of the velocity of the arrow 3.4 s after it leaves the bowstring? What are the x- and y-components of the displacement of the arrow during the 3.4 s interval?
Well, if I were an arrow, I would probably want to "dis-appear" too. But let's do some calculations first.
To find the x- and y-components of the velocity of the arrow, we can use some trigonometry. The x-component of the velocity (Vx) can be found by multiplying the initial velocity (20.1 m/s) by the cosine of the angle (60.3°):
Vx = 20.1 m/s * cos(60.3°)
And the y-component of the velocity (Vy) can be found by multiplying the initial velocity (20.1 m/s) by the sine of the angle (60.3°):
Vy = 20.1 m/s * sin(60.3°)
Now, let's calculate!
Vx = 20.1 m/s * cos(60.3°)
Vx ≈ 10.05 m/s
Vy = 20.1 m/s * sin(60.3°)
Vy ≈ 17.45 m/s
So, the x-component of the velocity is approximately 10.05 m/s, and the y-component is approximately 17.45 m/s.
Now, for the displacement of the arrow during the 3.4 second interval. We can use these components of velocity to calculate the x and y displacements separately.
The x-displacement (𝚫x) is equal to the x-component of velocity (Vx) multiplied by the time (3.4 s):
𝚫x = Vx * t
𝚫x = 10.05 m/s * 3.4 s
𝚫x ≈ 34.17 m
The y-displacement (𝚫y) is equal to the y-component of velocity (Vy) multiplied by the time (3.4 s), minus half the acceleration due to gravity (9.8 m/s²) multiplied by the square of the time (t²):
𝚫y = (Vy * t) - (0.5 * g * t²)
𝚫y = (17.45 m/s * 3.4 s) - (0.5 * 9.8 m/s² * (3.4 s)²)
𝚫y ≈ 44.23 m
So, the x-component of the displacement during the 3.4 second interval is approximately 34.17 m, and the y-component is approximately 44.23 m.
I hope that helps, and remember, arrows may go up, but they always come down... with style!
To solve this problem, we can use the equations of motion in projectile motion. Let's break the given information down step by step:
Step 1: Calculate the x-component of the velocity.
The x-component of the velocity remains constant throughout the motion.
Given that the arrow is shot at an angle of 60.3° above the horizontal, we can use trigonometry to find the x-component of the initial velocity (V₀x):
V₀x = V₀ * cos(θ)
V₀x = 20.1 m/s * cos(60.3°)
V₀x = 20.1 m/s * 0.5
V₀x = 10.05 m/s
Step 2: Calculate the y-component of the velocity.
The y-component of the velocity changes as the arrow moves upwards and then downwards due to gravity.
Given that the arrow is shot at an angle of 60.3° above the horizontal, we can use trigonometry to find the y-component of the initial velocity (V₀y):
V₀y = V₀ * sin(θ)
V₀y = 20.1 m/s * sin(60.3°)
V₀y = 20.1 m/s * 0.866
V₀y = 17.36 m/s
Step 3: Calculate the change in time.
The time interval given is t = 3.4 s.
Step 4: Calculate the y-component of the displacement (change in position).
The y-component of the displacement (Δy) can be calculated using the equation:
Δy = V₀y * t + (1/2) * g * t²
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Δy = 17.36 m/s * 3.4 s + (1/2) * 9.8 m/s² * (3.4 s)²
Δy = 59.10 m
Step 5: Calculate the x-component of the displacement.
The x-component of the displacement (Δx) can be calculated using the equation:
Δx = V₀x * t
Δx = 10.05 m/s * 3.4 s
Δx = 34.17 m
Step 6: Answer the question.
The x- and y-components of the velocity of the arrow 3.4 s after it leaves the bowstring are:
Vx = 10.05 m/s
Vy = 17.36 m/s
The x- and y-components of the displacement of the arrow during the 3.4 s interval are:
Δx = 34.17 m
Δy = 59.10 m
To find the x- and y-components of the velocity of the arrow, we can use trigonometry.
Given:
Initial velocity (v) = 20.1 m/s
Launch angle (θ) = 60.3°
The x-component of the velocity (v_x) can be found using the formula:
v_x = v * cos(θ)
The y-component of the velocity (v_y) can be found using the formula:
v_y = v * sin(θ)
Calculating the x- and y-components:
v_x = 20.1 m/s * cos(60.3°) ≈ 10.05 m/s
v_y = 20.1 m/s * sin(60.3°) ≈ 17.48 m/s
So, the x- and y-components of the velocity of the arrow 3.4 s after it leaves the bowstring are approximately 10.05 m/s and 17.48 m/s, respectively.
To find the x- and y-components of the displacement of the arrow during the 3.4 s interval, we can use the equations:
Δx = v_x * t
Δy = v_y * t - 0.5 * g * t^2
Given:
Time interval (t) = 3.4 s
Gravitational acceleration (g) = 9.8 m/s^2 (assuming no air resistance)
Calculating the x- and y-components of the displacement:
Δx = 10.05 m/s * 3.4 s ≈ 34.17 m
Δy = (17.48 m/s * 3.4 s) - (0.5 * 9.8 m/s^2 * (3.4 s)^2) ≈ 44.25 m
So, the x- and y-components of the displacement of the arrow during the 3.4 s interval are approximately 34.17 m and 44.25 m, respectively.