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

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To find the magnitude of the vector a-b, where a=3i-4j and b=4i-3j, subtract the corresponding components to get a-b = (-1i - 1j) and then apply the Pythagorean theorem to find the magnitude as sqrt((-1)^2 + (-1)^2) = sqrt(2). For the ball thrown at 26m/s at a 20-degree angle, resolve the initial velocity into horizontal and vertical components. The vertical component can be calculated using sin(20 degrees) multiplied by 26m/s, while the horizontal component uses cos(20 degrees). To determine the height after 1.4 seconds, establish the equation of motion considering the vertical velocity and gravitational deceleration, noting that if the time to reach maximum height is less than 1.4 seconds, the ball will be descending.
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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 ?
 
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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.
 
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