a 145g baseball is thrown up at 40 m/s. after traveling a distance of 50m in the air, the baseball is caught. the ball experienced air resistance of what magnitude?

Thrown straight up?

For the straight-up case, the ball must have travelled 25 m up and back down. A ball thrown at 40 m/s with no friction would have risen H = V^2/(2g) = 81.6 m

There is no way the height of a baseball thrown up in air will be reduced from 81.6 m to 25 m by friction.

This is a poorly conceived and inadequately explained question

If a baseball is thrown 30 m/s backwards from a truck moving 50 m/s, how fast will the ball strike the glove of a ground-based catcher?

To determine the magnitude of air resistance experienced by the baseball, we need to calculate the net force acting on it.

Assuming the air resistance is the only force acting on the baseball (neglecting gravitational force for now), the net force can be calculated using Newton's second law, which states that the net force (F_net) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a):

F_net = m * a

In this case, the baseball starts with an initial velocity (u) of 40 m/s and comes to a stop after traveling a distance (s) of 50m. The final velocity (v) will be 0 because the baseball is caught. Using the equation of motion (v^2 = u^2 + 2as), we can solve for acceleration:

0^2 = 40^2 + 2a * 50

1600 = 100a

a = 16 m/s^2

Now, we can plug the values of mass (m = 145g = 0.145kg) and acceleration (a = 16 m/s^2) into the equation for net force:

F_net = m * a
F_net = 0.145kg * 16 m/s^2
F_net = 2.32 N

Therefore, the magnitude of air resistance experienced by the baseball is approximately 2.32 Newtons (N).