A particle is projected up the incline such that it's component of velocity along the incline is 10 m/s. Time of flight is 2 sec and maximum height above the incline is 5m. Then velocity of projection will be.

A particle is projected up the incline such that it's component of vel

1.	A particle is projected up the incline such that it's component of velocity along the incline is 10 m/s. Time of flight is 2 sec and maximum height above the incline is 5m. Then velocity of projection will be.
Answer» v=10√2m/sUsing the given data in the formulae for projection up the inclined PLANE. (θ is angle of projectile with inclined plane, β is angle of inclined with horizontal) T=2sec=2vsinθgcosβ vcosθ=10 ---> (1) h=5=vsinθ2 × 1 vsinθ=10 from (i) and (II) v2=200 v=10√2m/sHOPE IT HELPS U !!

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A particle is projected up the incline such that it's component of velocity along the incline is 10 m/s. Time of flight is 2 sec and maximum height above the incline is 5m. Then velocity of projection will be.​

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A particle is projected up the incline such that it's component of velocity along the incline is 10 m/s. Time of flight is 2 sec and maximum height above the incline is 5m. Then velocity of projection will be.