AIPMT 2010 Physics and Chemistry Paper

Let the two balls meet after ts at distance x from the platform.
For the first ball
u=0,t=18s,g=10m/s2
Using h=ut+‌

1

2

gt2
∴‌‌x=‌

1

2

gt22....(i)
u×12+‌

1

2

×10×(12)2
For the second ball
u=u,t=12s,g=10m/s2
Using h=ut+‌

1

2

×10×102....(ii)
From equations (i) and (ii), we get
‌

1

2

×10×182=12u+‌

1

2

×10×(12)2
or 12u=‌

1

2

×10×[(18)2−(12)2]
=‌

1

2

×10×[(18+12)(18−12)]
12u=‌

1

2

×10×30×6
or ‌‌u=‌

1×10×30×6

2×12

=75m/s