1. A triangular prism has vertices at A(2, 0, 0), B(2, 1, 3), C(2, 2, 0), D(0, 0, 0), E(0, 1, 3), and F(0, 2, 0).
Which image point has the coordinates (1, 4, 3) after a translation using the vector 1, 2, 3?
2. What point represents a reflection of B over the xy-plane?
B'(?)
3. The vector has a magnitude of 6.1 inches and a direction of 55°. Find the magnitude of its veritcal component. (?) inches
4. Aaron kicked a soccer ball with an initial velocity of 39 feet per second at an angle of 44° with the horizontal.
After 0.9 second, how far has the ball traveled horizontally? (?)ft
After 1.5 seconds, how far has the ball traveled vertically? (?) ft
28. The vector has a magnitude of 5 inches and a direction of 32°. Find the magnitude of its vertical component. ?in.
3. Y = 6.1*sin55.
4. d = 39*cos44 * 0.9
Yo = 39*sin44 = 27.1 Ft/s. = Vertical component of initial velocity.
h = Yo*t + 0.5g*t^2.
h = 27.1*1.5 - 4.9*(1.5)^2
28. Y = 5*sin32
1. To perform a translation, we need to add the vector [1, 2, 3] to the coordinates of each point. Adding [1, 2, 3] to (1, 4, 3) gives us:
(1 + 1, 4 + 2, 3 + 3) = (2, 6, 6)
Therefore, the image point with coordinates (1, 4, 3) after the translation is (2, 6, 6).
2. Reflecting a point over the xy-plane simply means changing the sign of its z-coordinate, while keeping the x and y coordinates unchanged. In this case, the z-coordinate of point B is 3, so the reflection B' will have coordinates (2, 1, -3).
Therefore, the point that represents a reflection of B over the xy-plane is B'(2, 1, -3).
3. To find the vertical component of a vector with magnitude and direction, we can use trigonometry. We have the magnitude of the vector (6.1 inches) and the direction (55°), and we need to find the magnitude of the vertical component.
The vertical component can be found by multiplying the magnitude of the vector by the sine of the angle.
Vertical component = 6.1 inches * sin(55°) ≈ 4.964 inches
Therefore, the magnitude of the vertical component is approximately 4.964 inches.
4. We can use the equations of motion to calculate the horizontal and vertical distance traveled by the ball at different time intervals.
Given:
Initial velocity (u) = 39 feet per second
Launch angle (θ) = 44°
a) To find the horizontal distance traveled after 0.9 seconds, we need to find the horizontal component of the initial velocity and then multiply it by the time.
Horizontal component of velocity = u * cos(θ)
Horizontal distance traveled = (Horizontal component of velocity) * (Time)
Horizontal component of velocity = 39 ft/s * cos(44°) ≈ 27.82 ft/s
Horizontal distance traveled = 27.82 ft/s * 0.9 s ≈ 25.038 ft
After 0.9 seconds, the ball has traveled approximately 25.038 feet horizontally.
b) To find the vertical distance traveled after 1.5 seconds, we need to find the vertical component of the initial velocity and then multiply it by the time, and also consider the effect of gravity acting in the downward direction.
Vertical component of velocity = u * sin(θ)
Vertical distance traveled = (Vertical component of velocity) * (Time) - (0.5 * acceleration * Time^2)
Vertical component of velocity = 39 ft/s * sin(44°) ≈ 28.086 ft/s
Vertical distance traveled = (28.086 ft/s * 1.5 s) - (0.5 * 32.2 ft/s^2 * (1.5 s)^2) ≈ 13.693 ft
After 1.5 seconds, the ball has traveled approximately 13.693 feet vertically.
28. To find the vertical component of a vector with magnitude and direction, we can use trigonometry. We have the magnitude of the vector (5 inches) and the direction (32°), and we need to find the magnitude of the vertical component.
The vertical component can be found by multiplying the magnitude of the vector by the sine of the angle.
Vertical component = 5 inches * sin(32°) ≈ 2.638 inches
Therefore, the magnitude of the vertical component is approximately 2.638 inches.