Question

On the putting green, a golf ball travels at 1.4 m/s in a direction of 130°. The ball encounters wind blowing at 1.1 m/s in the direction of 50°. What is the true speed and direction of the ball? Round the speed to the thousandths place and direction to the nearest degree.

Answer 1

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:

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

Therefore the correct option is the second one.