A 0.42 kg football is thrown with a velocity of 17 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.017 S. What is the force exerted on the receiver? Answer in units of N

420 N

Explanation:

m = 0.42 Kg, u = 17 m/s, v = 0 m/s, t = 0.017 s

By first law of Newtons' laws of motion, the rate of change in momentum is force, F = m (v - u) / t

F = 0.42 x (0 - 17) / 0.017

F = - 420 N

Negative sign shows hat the force is resistive that means the ball finally comes to rest.