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# A particle is projected ...at that time its kinetic energy was KE1......and at the top point where vertical velocity is 0 its kinetic energy was KE2 ....... The ratio of KE2 : KE1 is 3:4 ........ find the difference in their velocities ......   Dear ankit let velocity at the begineeing ii u and angle of projection is θ. KE1 =1/2 mu2 KE2 =1/2 m(ucosθ)2 given KE2 : KE1 is 3:4 so cos2θ  =3/4 θ =30 difference in their velocity =u-ucosθ                                      =u-u√3/2                                      =u(1-√3/2)   sir but answer of this is u/2 ....

chaitanya pansare
23 Points
12 years ago

then might be the ques. is asking the difference in vertical velocities which comes out to be usinθ-0 i.e. u/2-0=u/2

148 Points
12 years ago

Dear Ankit Shukla

last answer was on the basis of difference in velocity magnitude.

I think you want difference in velocity vector

Initial velocity =ucosΘ i + usinΘ j

final velocity =ucosΘ i

difference in velocity =ucosΘ i + usinΘ j -ucosΘ i

=usinΘ j

=u sin30 j

=u/2 j

magnitude            =|u/2 j|

=u/2

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