a ball is thrown from the top of the tower in vertically upward direction.velocity at point h meter below the point of projection is twice of the velocity at apoint h meter abovethe point of projection.find the maximum height reached by the ball above the top of tower

Letu=Velocity of projectionVelocity at a point h below the point of projection = 2 × Velocity at a point h above the point of projectionu2+2gh=2u2-2ghSquare on both sides:u2+2gh=2u2-2ghu2+2gh=2u2-4ghu2=6ghHence,Maximum height isH=u22g   =6gh2g   =3h

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