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Blazing goIITian

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24 May 2012 11:16:26 IST

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Projectile Motion

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A particle is projected with velocity 'v' towards a vertical wall at distance 'x' from the point of projection. Then, find the maximum height by which the projectile will collide on the wall.

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Deevita Agarwal

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25 May 2012 10:54:59 IST

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After putting the angle in the trajectory equation, m getting y=(u^4-g^2x^2)/2gu^2

Deevita Agarwal

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25 May 2012 10:56:19 IST

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y=(u^4-g^2x^2)/2gu^2

Deevita Agarwal

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25 May 2012 10:57:07 IST

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y=(u^4-g^2x^2)/2gu^2

Deevita Agarwal

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25 May 2012 10:57:51 IST

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y=(u^4-g^2x^2)/2gu^2

umesh saini

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25 May 2012 15:08:10 IST

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one partical threw horizantaly from a bulding at the hight of 20 m. one other partical al

GOPAKUMAR

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2 Jun 2012 10:19:07 IST

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Deevita Agarwal

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4 Jun 2012 17:31:41 IST

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@gopakumar -- from where r u getting these two velocities, v and u??

GOPAKUMAR

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6 Jun 2012 07:06:19 IST

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SORRY IT'S NOT U BOTH ARE SAME "V"

Deevita Agarwal

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6 Jun 2012 21:06:31 IST

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yeah... its all right. :)

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