Answer
The magnitude of the horizontal component of $v$ is $25\sqrt{2}$
The magnitude of the vertical component of $v$ is $25\sqrt{2}$
Work Step by Step
We can find the magnitude of the horizontal component of $v$:
$v_x = v~cos~\theta$
$v_x = (50)~cos~45^{\circ}$
$v_x = (50)~(\frac{\sqrt{2}}{2})$
$v_x = 25\sqrt{2}$
The magnitude of the horizontal component of $v$ is $25\sqrt{2}$
We can find the magnitude of the vertical component of $v$:
$v_y = v~sin~\theta$
$v_y = (50)~sin~45^{\circ}$
$v_y = (50)~(\frac{\sqrt{2}}{2})$
$v_y = 25\sqrt{2}$
The magnitude of the vertical component of $v$ is $25\sqrt{2}$
