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The answer is (a) 0.2s (b) 0.404m

Since it was thrown with an angle why we need to consider the acceleration

T=2usintheta/g can we use that??

hey is time taken =1sec

sorry i mean 2sec ??

and yea shakir u can use dat formula u mention above

How in question it is given that theta=30

so wats the time of flight am i right?? i mean my ans is correct??

0.2s

but how?

i am also confused i did it with out taking any relative acceleration and velocity

still show me ur work wat u did

|dw:1338466097736:dw|

cany=*can

and yea shakir for the second option u have to cal range means u^2sin2theta/g

not getting that

i the distance 0.2sqrt3 ??

do u knw sin2theta means 2xsinthetaxcostheta

yes

|dw:1338466702370:dw|

but that is wrong

ok then wat abt range

but gautham dont u add lift's velocity too??

we dont have to ..think and tell

But not getting!!!!

Range should be 0.404 m somehow///

Any Idea Guys!

well i dont know how to do it in heena's way, although we can

wat abt Range!!!!!!!!!

im getting 0.044

so either time of flight isnt correct or shakir didnt type properly

no i am correct!

Confused!!

Time Wasting lets move to another question

ok lets do this from begining ok gautham we know lift i moving upwards means acceleration will g+a

@ heena wont it be a-g?

Try the second part :P

I TRIED THE SECON PART BUT I GOT IT AS 0.364 M BUT THE ANSWER IS 0.404M U CAN SEE MY WORK ABOVE

but y not @shivam_bhalla lift is not static u cant neglect that :O

@shivam_bhalla WAT ABT RANGE???

PLZZ SECOND THE SECOND PART OF THE QUESTION WE NEED TO FIND THE DISTANCE @shivam_bhalla

YES THEN THE DISTANCE IS ???????

HOW U=0.2?