A ball is thrown at 15.5 m/s at 42° above the horizontal. Someone located 30 m away along the line3 of the path starts to run just as the ball is thrown. How fast, and in which direction, must the catcher run to catch the ball at the level from which it was thrown?

Answer 1

The range of the projectile is

R(x) = 15.5^2 * sin(84°)/9.81 = 24.36 m

The height is
y(t) = 15.5 sin 42° t - 4.9t^2, so it takes 2.12 seconds to fall back to the initial height.

So, the runner has to run toward the thrower at (30-24.36)/2.12 = 2.66 m/s