Answer to Question #139584 in Physics for Audrianna White

Question

If a pitcher throws a fastball at 40 m/s. If there is a 12 m/s wind blowing from first base towards third base. I have included a diagram of what a baseball field looks like below.

What direction does the pitcher have to throw the pitch in order for it to be a cross directly over home plate and be a strike?

Answer 1

$$\theta=\arctan{\frac{v_w}{v}}\\\theta=\arctan{\frac{12}{40}}=16.7\degree$$