The assumption works.
Here's an example I often use when thinking of this. It may help.
In the diagram on the left, a ball resting on the floor.
In the diagram on the right, the same ball bouncing perfectly elastically up and down on the floor. It rises and falls a height h.
What is the average force the ball applies to the floor in the two cases?
(That is, averaged out over a long time.)
No problem with the left diagram. The force is constant and equal to mg.
What about the right one? The force on impact is quite high, but this only lasts for a short time. I don't even know what that time is.
However, I do know that it takes the ball a time T to fall and T to rise, giving a total of 2T between each bounce.
I have worked out a formula for T in the diagram.
I also know that the change in momentum is 2mv for this elastic bounce.
I have also worked out a formula for v in the diagram.
To find the average force of the bouncing ball on the floor over a time 2T, I need
average force = change in momentum / 2T = 2mv/2T
Note that if I want the average over a longer interval, say 4T, I need to remember that the momentum changes twice (2 bounces) in this time. So I actually get the same value 4mv/4T. It doesn't matter how long an interval I take, the average force will be the same. This is the force the bouncing ball applies to the ground averaged over a long time. (It is the value the force would be if it was constant.)
What do you expect this force to be?
If you do the maths and find 2mv/2T by substituting in the values for v and T, you will find the answer. It should not surprise you.
By the way, I've seen this question (find the average force of the bouncing ball on the ground) asked at A level. It's worth making a note of.