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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 mu 2 KE2 =1/2 m(ucosθ) 2 given KE2 : KE1 is 3:4 so cos 2 θ =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 ....
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
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 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards, Askiitians Experts Badiuddin
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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