Question #0e11f

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Question #0e11f

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Answer 1 · 2016-12-27T00:52:23

See below.

Explanation:

Here is a basic diagram for the launch of a projectile, where $θ$ $theta$ is the launch angle, $v$ $v$ is the launch velocity of the projectile, and $v_{x}$ $v_x$ and $v_{y}$ $v_y$ are the horizontal and vertical components of the launch velocity, respectively.

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We can remember the most basic trig. functions for a right triangle by "SOH CAH TOA," where $sin (θ)$ $sin(theta)$ is equal to the ratio of the opposite side over the hypotenuse, $cos (θ)$ $cos(theta)$ is equal to the ratio of the adjacent side over the hypotenuse, and $tan (θ)$ $tan(theta)$ is equal to the ratio of the opposite side over the adjacent side. For our triangle above,

$sin (θ) = \frac{y}{r}$ $sin(theta)=y/r$

$cos (θ) = \frac{x}{r}$ $cos(theta)=x/r$

$tan (θ) = \frac{y}{x}$ $tan(theta)=y/x$

Where $x$ $x$ is the adjacent side, $y$ $y$ is the opposite side, and $r$ $r$ is the hypotenuse.

If we wanted to find the vertical component of velocity, $v_{y}$ $v_y$ , given the overall velocity, $v$ $v$ , we would use $sin (θ)$ $sin(theta)$ :

$sin (θ) = \frac{v_{y}}{v}$ $sin(theta)=(v_y)/v$

Solving for $v_{y}$ $v_y$ ,

$v_{y} = v sin (θ)$ $v_y=vsin(theta)$

If we wanted to find the vertical component of velocity, $v_{x}$ $v_x$ , given the overall velocity, $v$ $v$ , we would use $cos (θ)$ $cos(theta)$ :

$cos (θ) = \frac{v_{x}}{v}$ $cos(theta)=(v_x)/v$

Solving for $v_{x}$ $v_x$ ,

$\Rightarrow v_{x} = v cos (θ)$ $=>v_x=vcos(theta)$

If given the horizontal (vertical) component, we can find the vertical (horizontal) component using $tan (θ)$ $tan(theta)$ :

$tan (θ) = \frac{v_{y}}{v_{x}}$ $tan(theta)=(v_y)/(v_x)$

$v_{y} = v_{x} tan (θ)$ $v_y=v_xtan(theta)$

$v_{x} = \frac{v_{y}}{tan θ}$ $v_x=(v_y)/(tantheta)$

These equations are only valid when the horizontal component is adjacent to the launch angle.

If you are dealing with a physics problem that puts your components in a different quadrant, you may have the horizontal component opposite the angle and the vertical component adjacent to the angle. If this is the case, simply flip the equations. For example, you would find the horizontal component using $sin (θ)$ $sin(theta)$ and the vertical component using $cos (θ)$ $cos(theta)$ . In just above every basic projectile motion problem, however, you will find that the above equations hold.

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Question #0e11f

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