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Two particles thrown up simultaneously with velocity of 30m/s ,one from vertically and another at 45 degree with horizon.Fi d out the distance between them at t=2.5 s.

Two particles thrown up simultaneously with velocity of 30m/s ,one from vertically and another at 45 degree with horizon.Fi d out the distance between them at t=2.5 s.

Grade:12th pass

1 Answers

Mayank Ranka
askIITians Faculty 277 Points
2 years ago
Hi there,
Here’s your answer

We can use the concept of relative velocity
Let the first particle be at rest, so it remains at origin
but we have to add -30m/s to its velocity for it to come at rest.
Hence we must add the same to the vertical component of the velocity of the second particle
Hence the vertical component of initial velocity becomes 30sin45-30.
Now there is avertical acceleration of g downwards..
Hence y= ut +1/2at² = -8.79t-5t² =-13.185-11.25=-24.435....
Now horizontal velocity is constant
Hence x=ut=30cos45*t=2.115
Now the distance between origin and the point would be v(x²+y²) =v(173.844+4.473)=v178.317=13.35m

Thanks
Regards
Ankit Gupta

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