Average acceleration of a falling ball on contact with floor

[bryansonex]

Homework Statement

A ball falls on the surface from 10m height and rebounds to 2.5m.If duration of contact with floor is 0.01 sec,then average acceleration during contact is-
g = 10m/s^2

Homework Equations

<br /> \alpha = \frac{v_2 + v_1}{t}<br />

The Attempt at a Solution

Now here
<br /> v_{1} = \sqrt{2gh_{1}<br />

<br /> v_{1} = \sqrt{50}<br />

and

<br /> v_{2} = \sqrt{2gh_{2}<br />

<br /> v_{2} = \sqrt{200}<br />

<br /> \alpha = \frac{\sqrt{200+50}}{0.01}<br />


After solving all of this,I get the answer the answer as 1581.14 m/s^2,but the answer in the book is 2100 m/s^2.Am i doing anything wrong over here?

 

[nasu]

You'll get the answer if you do the calculation more carefully.
An estimate, without calculator:
sqrt(50) approx. 7
sqrt(200) is 10sqrt(2)= 14 (approx.)
7+14=21
21/0.01 = 2100

Your mistake: sqrt(50)+sqrt(200) is NOT sqrt(50+200)

 

[bryansonex]

ohh...thanks nasu