(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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 - >[tex]\frac{2h}{t_{2}^2-t_{1}^2}[/tex]

Was wondering how to get that expression,I've done it b4 but forgot I'm afraid,pls guide?

2. Relevant 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.

3. The attempt at a solution

Well,I've [tex]\frac{h+x}{t_{2}}[/tex] as v

[tex]\frac{x}{t_{1}}[/tex] as u

So I do (v-u)/t but I get something far from what the correct answer is ie

[tex]\frac{t_{1}(h+x)-xt_{2}}{t_{2}t_{1}(t_{2}-t_{1})}[/tex]

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# Homework Help: Simple acceleration puzzle I need help with,I forgot how to do.

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