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Pejman · Answer 1 · 2023-12-21T09:43:32+0000

In order to calculate the true speed and direction of the ball, let's calculate the horizontal and vertical directions of the ball speed (b) and of the wind speed (w):

$\begin{gathered} b_x=b\cdot\cos(\theta)\\ \\ b_x=1.4\cdot\cos(130°)\\ \\ b_x=1.4\cdot(-0.6428)\\ \\ b_x=-0.900\text{ m/s}\\ \\ \\ \\ b_y=b\cdot\sin(\theta)\\ \\ b_y=1.4\operatorname{\cdot}\sin(130°)\\ \\ b_y=1.4\operatorname{\cdot}0.766\\ \\ b_y=1.072\text{ m/s} \end{gathered}$
$\begin{gathered} w_x=w\cdot\cos(\theta)\\ \\ w_x=1.1\operatorname{\cdot}\cos(50°)\\ \\ w_x=1.1\operatorname{\cdot}0.6428\\ \\ w_x=0.707\text{ m/s}\\ \\ \\ \\ w_y=w\operatorname{\cdot}\sin(\theta)\\ \\ w_y=1.1\operatorname{\cdot}\sin(50°)\\ \\ w_y=1.1\operatorname{\cdot}0.766\\ \\ w_y=0.843\text{ m/s} \end{gathered}$

Now, let's add the horizontal components together and the vertical components together:

$\begin{gathered} v_x=b_x+w_x=-0.9+0.707=-0.193\text{ m/s}\\ \\ v_y=b_y+w_y=1.072+0.843=1.915\text{ m/s} \end{gathered}$

To calculate the magnitude and direction of this resultant speed, we can use the formulas below:

$\begin{gathered} v=√(v_x^2+v_y^2)\\ \\ v=√((-0.193)^2+(1.915)^2)\\ \\ v=1.925\text{ m/s}\\ \\ \\ \\ \theta=\tan^(-1)((v_y)/(v_x))\\ \\ \theta=\tan^(-1)((1.925)/(-0.193))\\ \\ \theta=96° \end{gathered}$

Therefore the correct option is the second one.

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