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

1. May 30, 2007

### inv

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 - >$$\frac{2h}{t_{2}^2-t_{1}^2}$$
Was wondering how to get that expression,I've done it b4 but forgot I'm afraid,pls guide?

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 $$\frac{h+x}{t_{2}}$$ 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
$$\frac{t_{1}(h+x)-xt_{2}}{t_{2}t_{1}(t_{2}-t_{1})}$$

Last edited: May 30, 2007
2. May 30, 2007

### andrevdh

Use

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

3. May 30, 2007