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Question: How do I find the vertical component of a vector?...

How do I find the vertical component of a vector?

Explanation

Solution

This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. To solve this problem we need to know the formula to find the vertical component. Also, we need to know the basic form of vector. To solve this question we need to know the formulae related to finding tan⁡θ\tan \theta and magnitude of the vector.

Complete step by step solution:
In this question, we would explain how to find the vertical component of a vector. For this, we would know the basic form of vector.
The basic form of a vector is shown below,
Vector=ai→+bj→= a\overrightarrow i + b\overrightarrow j
The formula for finding the vertical component is shown below,
Vy=∣A∣sin⁡θ→(1){V_y} = \left| A \right|\sin \theta \to \left( 1 \right)
Here θ\theta is the angle and ∣A∣\left| A \right| is the magnitude of the vector AA. To find the magnitude of the vector we use the following formula,
∣A∣=a2+b2\left| A \right| = \sqrt {{a^2} + {b^2}}
Next, we would find the value of θ\theta . So, we use the following formula
tan⁡θ=ba\tan \theta = \dfrac{b}{a}
When we move the term tan⁡\tan from the left side to the right side of the equation it converts into arctan⁡\arctan .
So, we get
θ=arctan⁡(ba)\theta = \arctan \left( {\dfrac{b}{a}} \right)
Here, aa and bb are the coefficients ofi→\overrightarrow i and $$\overrightarrow j

**So, by using the $$\theta $$ value we can easily find the value of $$\sin \theta $$. By using $$\left| A \right|$$ value and $$\sin \theta $$ value we can easily find the vertical component of the vector $${V_y}$$.** **Note:** This question involves the operation of addition/ subtraction/ multiplication/ division. Note that the denominator value would not be equal to zero. Note that the value of $${i^2}$$ and $${j^2}$$ is equal to$$ - 1$$. Remember that the horizontal component of a vector is mentioned as $${V_X}$$ and the vertical component of a vector is mentioned as$${V_y}$$. Note that when we move $$\tan $$ from left to right or right to left side of the equation it converts into$$\arctan $$.