If you have a 60 g egg dropped from a height of 10 m with cushioning material that delays the impact by 2 seconds what was the force exerted on the egg by the floor?

The potential energy of the egg when it begins to fall is equal to the kinetic energy of the egg just before touching the ground:
 m * g * h = (1/2) * m * v ^ 2
 Clearing the speed:
 v = root (2 * h * g)
 Substituting the values
 v = root (2 * (10) * (9.8)) = 14
 Then, by definition, we have
 F * deltat = m * deltav
 Clearing F
 F = (m * deltav) / (deltat)
 Substituting the values
 F = ((0.060) * (14)) / (2) = 0.42
 answer
 the force exerted on the egg by the floor was 0.42N

4. 50x10= 500m/s 5. 52m/s