Question

PE: A golf ball is hit 42 m/s due north, but a strong wind is blowing to the east at 12 m/s. What is the resulting velocity of the golf ball including magnitude & direction?

Answer 1

43.68 m/s with a direction of 74.05°

Explanation:

We make the sketch of the situation, just like this:

We can calculate the magnitude of the resulting velocity just like this (By means of the Pythagorean theorem since a right triangle is formed):

`\begin{gathered} h^2=a^2+b^2\rightarrow h=√(a^2+b^2) \\ \\ V_(R)=42^(2)+12^(2) \\ \\ V_R=√(1764+144) \\ \\ V_R=√(1908) \\ \\ V_R=43.68\text{ m/s} \end{gathered}`

While the direction is calculated as follows (applying the tangent trigonometric ratio):

`\begin{gathered} \tan\theta=(V_(north))/(V_(east)) \\ \\ \text{ We replacing} \\ \\ \tan\theta=(42)/(12) \\ \\ \theta=\tan^(-1)\left((42)/(12)\right)\: \\ \\ \:θ=74.05\degree \end{gathered}`

Which means that the resulting velocity is 43.68 m/s with a direction of 74.05°