PHY MCQ helpppp!!!!

[budhanilkantha]

7 Two markers M1 and M2 are set up a vertical distance h apart.
A steel ball is released at time zero from a point a distance x above M1. The ball reaches M1 at
time t1 and reaches M2 at time t2. The acceleration of the ball is constant.
Which expression gives the acceleration of the ball?

(refer oct/nov 2005 q.no 7 )

 

[diwash]

dude...
first of all ...the height travelled in first t1 time is..
x = o + 1/2 a*(t1)^2....(as the body starts from rest)...............................eqn (I)
and from the rest position in time t2 the ball fall a height of (h+x)..
so
h+x = 0+ 1/2 a*(t2)^2....................................eqn (II)
and we know the change in height from t1 to t2 is h ..so h should be given by subtracting the right hand expression of both the eqns...
so
h = [1/2 a*(t2)^2 ] -[1/2 a*(t1)^2]
h= 1/2 a* [ (t2)^2 - (t1)^2]
...now solve the eqn and u will get a = 2h/[ (t2)^2 -(t1)^2]....so answer is D.

 

[TaffsAsLevel]

dude...
first of all ...the height travelled in first t1 time is..
x = o + 1/2 a*(t1)^2....(as the body starts from rest)...............................eqn (I)
and from the rest position in time t2 the ball fall a height of (h+x)..
so
h+x = 0+ 1/2 a*(t2)^2....................................eqn (II)
and we know the change in height from t1 to t2 is h ..so h should be given by subtracting the right hand expression of both the eqns...
so
h = [1/2 a*(t2)^2 ] -[1/2 a*(t1)^2]
h= 1/2 a* [ (t2)^2 - (t1)^2]
...now solve the eqn and u will get a = 2h/[ (t2)^2 -(t1)^2]....so answer is D.

Hi but why isn't that 2h/(t2-t1)^2 because that is the time between h anyways? Please help