Using Newton’s 2nd law is the long way of solving this problem, but here goes.
First, we have to find the deceleration of the ball, coming to rest in the catcher’s mitt. The initial velocity of the baseball is 30 m/S, and the final velocity of the ball is 0.0 m/s; Vave = [V0 + Vf ]/2 = 15.0 m/s. During its deceleration, the ball has an average velocity of 15 m/S. We need this information to determine how long the deceleration takes. Remember: Speed = distance (Δ x) / time (seconds). We know that Δ x = 10 cm (the distance the mitt moves), and the average velocity of the ball during its deceleration = 15 m/S. We can rearrange the equation to solve for time: t=d/S. 0.1m/15.0 m/S = 6.67mS; so the ball takes 6.67 milliseconds to totally decelerate to 0.0 m/S. Acceleration = Δ V / Δ t. Δ V = -30.0 m/S, and t = 6.67 mS. (-30.0 m/s) / 6.67 mS = -4500 m/S^2. We know the mass of the ball is 0.1 kg. Plugging in the numbers to the famous equation F=ma, we get 0.1 kg * -4500 m/S^2= -450 Newtons of force. <-- you can disregard the negative sign; whether it’s positive or negative depends on your frame of reference.