Answer to Question #22833 in Mechanics | Relativity for matt
A 0.43 kg football is thrown with a velocity
of 12 m/s to the right. A stationary receiver
catches the ball and brings it to rest in 0.023
What is the force exerted on the receiver?
Answer in units of N
F=m*a, second newton's law. F=0.43*a -& exerted force a=v/t=12*0.023 - we find the acceleration, that stopped the ball while force was exerted So F=0.43*12/0.023=224 N - we put in numbers to find the answer