Question

a 0.30kg ball encounters a catchers hand while moving at the rate of 40m/s. the catcher draws his hand back so that the ball comes to a stop in a distance of 50cm. how strong a force did the ball exert on the catchers hand?
i got 29.43N

Answer 1

Wait never mind i got 480N

Answer 2

V^2 = Vo^2 + 2a*d = 0.
40^2 + 2a*0.05 = 0,
1600 + 0.1a = 0,
a = -16000 m/s^2.

F = M*a = 0.3 * (-16,000) = -4800 N.
The negative sign means the force opposes the mottion.