A0.5-kg ball moving at 5 m/s strikes a wall and rebounds in the opposite direction with a speed of 2 m/s. if the impulse occurs for a time duration of 0.01 s, then the average force (magnitude only) acting upon the ball is newtons.

Given that:

                       mass of the ball (m) = 0.5 Kg ,

                    ball strikes the wall (v₁) = 5 m/s ,

rebounds in opposite direction (v₂) = 2 m/s,

                                time duration (t) = 0.01 s,

        Determine the force (F) = ?

We know that from Newton's II law,

                                F = m. a  Newtons  

                                  (velocity acting in opposite direction, so a = ( (v₁ + v₂)/t

                                   = m × (v₁ + v₂)/t

                                   = 0.5 × (5 + 2)/0.01

                                  = 350 N

The force acting up on the ball is 350 N

i think the answer is a, hope this : )