pulley system: show speed of each particle is 3root3g/8 m/s

[markosheehan]

Two particles of masses 3 kg and 5 kg are connected by a light inextensible string, of length 4 m, passing over a light smooth peg of negligible radius. The 5 kg mass rests on a smooth horizontal table. The peg is 2.5 m directly above the 5 kg mass. The 3 kg mass is held next to the peg and is allowed to fall vertically a distance 1.5 m before the string becomes taut.
Show that when the string becomes taut the speed of each particle is 3root3g all over 8 m/s
i tried working this out by looking at the 3kg weight and using the equation v=u+2as
u=0 a=g s=1.5 however working this out gives me an answer of root3g this is not the desired answer

 

[stapel]

Two particles of masses 3 kg and 5 kg are connected by a light inextensible string, of length 4 m, passing over a light smooth peg of negligible radius. The 5 kg mass rests on a smooth horizontal table. The peg is 2.5 m directly above the 5 kg mass. The 3 kg mass is held next to the peg and is allowed to fall vertically a distance 1.5 m before the string becomes taut.
Show that when the string becomes taut the speed of each particle is 3root3g all over 8 m/s
i tried working this out by looking at the 3kg weight and using the equation v=u+2as
u=0 a=g s=1.5 however working this out gives me an answer of root3g this is not the desired answer

Please reply showing your work, and explaining things such as "the equation v = u + 2as" (and for what each variable stands). Thank you!

 

[markosheehan]

its ok i worked out