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How have u concluded it to b sin inverse x ?

Hi every1...... Here in this thread u ve 2 specify whether u ve got spectacles or not. Lemme start ...... Nopes m not a specky..

Hi...I've seen d soln provded by magiclko in D Mukherjee. So chck it out ther in d book for furthr info

Can u explain why x is zero ??

I wsa not wrong. I did not look at the ans properly. THanks sir .

Yes its MCQ by D Mukherjee. Solution given is :- v = horizontal comp velocity, T = total tome of flight , vT = 2 km

When the shel breaks the part which falls verticaly doesnt have any horizontal vel & hence the other part acquires horizontal vel 2v , by momentm consrvation. Also its remainng time of flight wil b T/2. It wil travel a further distance 2v x T/2 = vT = 2 km .Hene it wil land 3 km from canon. I have given what the book has given .

Though what u r sayin is not fully right but thanks I ve got a gud clue from this.I ve solved it.

Hi Pl explain why not 2 times?

A cannon shel lands 2 km away from the cannon . A 2nd shell , fired identically , breaks into 2 equal parts at the highest point . 1 falls vertically.How far from the cannon wil the other land?

The solution I have has done it with momentum conservation. Ans given is 2 km. Can I use the following:- The 2 parts shud have their COM at 2 km. 2=m+m[2+x]/2m. By this I m gettin x = 1 , so 3km. Pl explain wher m goin wrong.

A spring of weight W and force constant k is suspended in a horizontal position by two light strings attached to its two ends. Each string makes an angle theta with the vertical . What will be the extension of the spring??

The solution I have has given the force in horizontal position Tsin theta=kx

I want to ask that the Tsin theta acts in left and right directions .So why should not it get cancel?