A 4.0 kg ball is traveling at 3.0 m/s and strikes a wall. The ball bounces off the wall with a velocity of 4.0 m/s in the opposite direction. If the ballis in contact with the wall for 0.10 seconds, then the magnitude of the average force on the ball during contact is

a. 230 N.
b. 400 N.
c. 360 N.
d. 320N,

280 N

Explanation:

Applying Newton's third second law of motion,

F = m(v-u)/t Equation 1

Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.

Note: Let the direction of the initial velocity of the ball be positive

Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s

Substitute into equation 1

F = 4(-4-3)/0.1

F = 4(-7)/0.1

F = -28/0.1

F = -280 N.

Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball

the answer is a. hope this .

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr