A 5 kg ball approaches a wall at a speed of 4 m/s. It then bounces off the wall in the opposite direction at the same speed. What is the magnitude of the average force exerted on the ball if it is in contact with the wall for 0.1 s?

Average force =  2 kg m/s²

Explanation:

Given

mass = 5 kg

initial velocity = 0 m/s

final velocity = 4 m/s

time = 0.1 sec

Find

average force = ?

Formula

Average force = m (final velocity - initial velocity)/t₂- t₁

                        = (5kg)(0 m/s - 4 m/s)/0 - 0.1

                        = 5 kg (- 4 m/s)/(- 0.1 sec)

                         = 2 kg m/s²

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

tdhnjgrgf

explanation: