Change of the momentum of the ball dropped on the floor

A ball of mass 𝑚 = 8.0 ∙ 10−2 𝑘𝑔 starts from rest and falls vertically downward from a height of 3.0 𝑚. After colliding with the ground, it bounces up to a height of 2.0 𝑚.
The collision takes place over a time interval of ∆𝑡 = 5.0 ∙ 10−3 𝑠. Calculate:
1) the momentum of the ball immediately before and immediately after the collision
2) average force exerted by the ground on the ball
3) impulse imparted to the ball

[MrJopi]

You forget to take into account the direction of the moving ball: dp is in this case the difference between -0, 61 kgm/s and +0, 50 kgm/s = +1, 11 kgm/s, resulting in a (average) net force of 222 N upwards during impact. Also: The average force exerted by the ground will then be: F_net + Fg = 222 + 0, 08*9, 8 = 223 N, because F_net = F_ground - Fg

[tayyabaishtiaq]

how is he writing backwards SO SMOOTHLY ??!! am i missing something or

[ppmenotsmall]

May I ask what is the change in kinetic energy during the collision?

[MoizAsimagar]

Please explain how is the momentum conserved?

[pradeepojha]

A ball of mass kg, .10 starting from rest, falls a height of 4.0 m and then collides with the ground. After the collision, the ball bounces up to a height of 2.0 m. The collision with the ground takes place over a time .s100.43 Determine (i) the momentum of the ball immediately before the collision and immediately after the collision and (ii) the average force exerted by the ground on the ball. Take .ms 0.102g ...solution for this

[aleksanderamid]

Haha, this is wrong due to the direction errors you make in the momentum and velocity xd