Einstein solved this when he was 12...?

Question

My answer was ~1000...but im not sure about the reasoning.
Could you please help me with this problem?

"A hard rubber ball, realeased at chest height, falls to the pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattned. Suppose the maximum depth of the dent is on the order of 1cm. Compute an order of magnitude estimate for the maximum acceleration of the ball while it is in contact with the pavement"

Answer 1

Mass of ball: m

It falls a distance of about 1 meter ( = h), so it has speed given by:
mv^2/2 = mg*h
v = sqrt(2g*h)

All the motion has to be wiped out within 1 cm ( = d=: So the work done by the force is:
W = F*d = ma*d, and this must equal the KE:
mv^2/s = KE = F*d = ma*d, so:
a = (v^2/2)/d = gh/d

So a = g*(h/d) = g*(1 m / 1 cm) = 9.8 * 100 = 980.

(Tricky point: the "a" calculated here is the acceleration due to the "net" force. The "normal" force, which is actually exerted by the concrete, has in addition another mg to counter-act the weight of gravity. The normal force is what you would feel if you bounced the ball off your foot instead of the concrete.)

Answer 2

Well, that's pretty easy. By using the acceleration due to gravity, you'll easily figure out how fast the ball was traveling. You know it stopped in 1 cm. You don't know if the deceleration was constant, but heck, they're only asking for an order of magnitude.

Answer 3

I suppose nothing...........