Question

Vector A of magnitude 5.0 units has a horizontal component of 2.0 units. Find the angle that vector A makes with the horizontal and determine the vertical component of vector A.

Answer 1

Answer:

Let vector A = 2 i^ + b j^

it's magnitude =

$\sqrt{ {2}^{2} + {b}^{2} } = 5$

$4 + {b}^{2} = 25$

${b}^{2} = 25 - 4 = 21$

$b = \sqrt{21} = y \: component$

so vector A = 2 i^ + √21 j^

Let required angle = c

$c = {tan}^{ - 1} ( \frac{ \sqrt{21} }{2} ) \: \: ans$

Answer 2

Explanation:

Given:

Magnitude of vector $|A|=5 u$

Horizontal component $=2 u$

To find: Angle that vector $A$ makes with the horizontal and vertical component of vector $A$

Calculation:

Any vector has two components, horizontal and vertical.

Let α be the angle made by vector$A$ with the horizontal.

Horizontal component $A_{x}$ is given as $=|A| cos\alpha$ where α is the angle made by the vector with horizontal component.

So,

$A_{x}=2 \\A_{x} =|A| cos \alpha \\2 =5cos \alpha \\\alpha =cos^{-1} (\frac{2}{5})$

Angle made by  vector $A$ with the horizontal is $\alpha =cos^{-1} (\frac{2}{5} )$

Vertical component of vector $A$,

$A_{y}=|A| sin\alpha \\A_{y}=5sin(cos^{-1} (\frac{2}{5} ))$

Form trigonometric identities:

$sin(cos^{-1} x)=\sqrt{1-x^{2} }$

So,

$A_{y}=5sin(cos^{-1} (\frac{2}{5} ))\\A_{y}=5\sqrt{1-(\frac{2}{5}) ^{2} } \\A_{y}=\sqrt{21}$

Therefore, $A_{y}=\sqrt{21}$

Answer:

Angle made by the vector with horizontal component $\alpha =cos^{-1} (\frac{2}{5} )$

Vertical component $A_{y}=\sqrt{21}$