Simple acceleration puzzle I with,I forgot how to do.

[inv]

Homework Statement

Two markers M1 & M2're set up a vertical distance h apart.

When a steel ball is released from rest from a pt a distance x above M1,it's found that the ball takes time t1 to reach M1 and time t2 to reach M2.
Which expression gives the acceleration of the ball?
A.Wrong,B.Wrong,C.Wrong,D.Correct - >\frac{2h}{t_{2}^2-t_{1}^2}
Was wondering how to get that expression,I've done it b4 but forgot I'm afraid,pls guide?

Homework Equations

Acceleration is the rate of change of velocity,ie (v-u)/t where v is final velocity,u is initial velocity,t is time taken for the change in velocity to occur.

The Attempt at a Solution

Well,I've \frac{h+x}{t_{2}} as v
\frac{x}{t_{1}} as u
So I do (v-u)/t but I get something far from what the correct answer is ie
\frac{t_{1}(h+x)-xt_{2}}{t_{2}t_{1}(t_{2}-t_{1})}

 

[andrevdh]

Use

s = ut + \frac{1}{2}gt^2

 

[inv]

Solved*!Tq for answering!