Solving Vector Applications: a-b Magnitude & Ball's Height

[Huskies213]

Can anyone explain what to do for these 2 ?

1.) If a=3i-4j, and b=4i-3j, what is the magnitude of the vector a-b?


2.) A boy throws a ball at an initial velocity of 26m/s at an angle of 20 degrees above the horizontal. How high above the projection point is the ball after 1.4 seconds ?

 

[Astronuc]

For the vector problem, i,j represent usually represent mutually orthogonal (perpendicular) vectors, e.g. in x,y-directions.

To add two vectors, one adds corresponding components.

let a = a₁i + a₂j, and b = b₁i + b₂j, so a + b = (a₁+b₁)i + (a₂+b₂)j. The subtraction is just the additive inverse, i.e. replace + with -.

The magnitude of a is just sqrt(a₁²+a₂²). Remember the formula for the length of the hypotenuse of a right triangle - Pythagorean theorem.

In question 2, one has to resolve the vector given by "26m/s at an angle of 20 degrees" into horizontal and vertical components. The ball travels vertically with some velocity component, but decelerates due to gravity, reaches a maximum at some time T, and returns to the same initial elevation at time 2T (i.e. T up and T down - neglecting air resistance).

Using the vertical velocity component, one can establish the equation of motion as a function of time to determine where the ball is at 1.4 s. If T (time to max height) < 1.4 s, then the ball is falling back from its maximum altitude.

 

[kyle8921]

1. To solve for the magnitude of the vector a-b, we can use the Pythagorean theorem. The magnitude of a vector is the length of the vector, which can be found by taking the square root of the sum of the squares of its components. In this case, we have a horizontal component of 3 and a vertical component of -4 for vector a, and a horizontal component of 4 and a vertical component of -3 for vector b. So, the magnitude of the vector a-b can be calculated as follows:

|a-b| = √[(3-4)^2 + (-4-(-3))^2]
= √[(-1)^2 + (-1)^2]
= √2
≈ 1.41

Therefore, the magnitude of the vector a-b is approximately 1.41 units.

2. To solve for the height of the ball after 1.4 seconds, we can use the equations of motion for projectile motion. The vertical displacement of the ball can be calculated as follows:

y = y0 + v0sinθt - 1/2gt^2

Where:
y = vertical displacement (height)
y0 = initial height (0 in this case)
v0 = initial velocity (26m/s)
θ = angle of projection (20 degrees)
t = time (1.4 seconds)
g = acceleration due to gravity (9.8m/s^2)

Substituting the given values, we get:

y = 0 + (26m/s)(sin20°)(1.4s) - 1/2(9.8m/s^2)(1.4s)^2
= 8.6m - 9.66m
≈ -1.06m

This means that after 1.4 seconds, the ball is 1.06 meters below its initial projection point. To find the height above the projection point, we need to take into account the initial height of the ball, which is 0 in this case. Therefore, the height above the projection point after 1.4 seconds is 1.06 meters.