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Sagot :
In order to find the direction of the initial velocity, first let's calculate the change in velocity caused by the acceleration:
[tex]\begin{gathered} F=m\cdot a\\ \\ 2655=15.5\cdot a\\ \\ a=171.29\text{ m/s^^b2}\\ \\ \\ \\ \Delta V=a\cdot t\\ \\ \Delta V=171.29\cdot0.05\\ \\ \Delta V=8.5645\text{ m/s} \end{gathered}[/tex]Now, let's calculate the horizontal and vertical components of the final velocity.
The direction is N45W, therefore the angle is 90° + 45° = 135°:
[tex]\begin{gathered} V_x=V\cdot\cos135°=45\cdot(-\frac{\sqrt{2}}{2})=-31.82\text{ m/s}\\ \\ V_y=V\cdot\sin135°=45\cdot\frac{\sqrt{2}}{2}=31.82\text{ m/s} \end{gathered}[/tex]The change in the velocity occurred in north direction, so let's calculate the initial vertical velocity:
[tex]V_{iy}=V_y-\Delta V=31.82-8.5645=23.2555\text{ m/s}[/tex]Now, to calculate the direction, we can use the arc tangent below:
[tex]\theta=\tan^{-1}(\frac{V_{iy}}{V_{ix}})=\tan^{-1}(\frac{23.2555}{-31.82})=-36.16°[/tex]Since the direction is between north and west, we need to add 180° to the result, so the angle will be 143.84°.
This angle is equal to the direction N 53.84° W.
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