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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 ....

```
8 years ago

chaitanya pansare
23 Points
```										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
```
8 years ago
147 Points
```										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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```
8 years ago
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