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|dw:1360985504635:dw|

what i did is |dw:1360985802827:dw|

for h the total distance id 1m+.2m (compressed distance )

sin37(1.2)=h

yes you are proceeding correctly

if i solve it i got v=4.20m/s

is this right

for (a)

no this will be the answer to part b i.e. the canister has reached the ground

(b) i got |dw:1360986031295:dw|

you have calculated h corresponding to the lower end of the incline

it doest not matter right because i got height so the box is top so height works

|dw:1360986099560:dw|

|dw:1360986365438:dw|

u mean that

yes the black line shows the heigth 1.2 sin37

shouldnt that be 1sin 37 because the spring not compressed

for part a energy conservation will be |dw:1360986436055:dw|

|dw:1360986652778:dw|

u mean that right

okay for part b u need to use h=1.2sin37 and should we consider the part (a) speed as initial speed

|dw:1360986763295:dw|

you may do that
here h = 1*sin37

since the body has already covered 0.2m in part a as it loses contact with the spring

for part b u mean

okay so h=1 sin 37 and i can use above equation right with initial speed from part (a)

yes for part b

okay thx very much