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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 ....

Ankit Shukla , 15 Years ago

Grade

2 Answers

chaitanya pansare

Last Activity: 15 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

Badiuddin askIITians.ismu Expert

Last Activity: 15 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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