one second after projection, a stone moves at an angle 45° with horizontal. two second after projection, it moves horizontally. its angle of projection is tan^-1 (2). Then what is its initial velocity.

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Vikas TU · Answer

General eqn. at time gt in x and y direction respectively are = > Vx = ucos(theta) Vy = usin(theta) – gt At t = 2 s, Vy = 0 usin(theta) – 2g = 0 u = 2g/sin(theta) = > 2g/(2/root5) { given theta = tan^-1 (2)} u = g*root(5) m/s.

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one second after projection, a stone moves at an angle 45° with horizontal. two second after projection, it moves horizontally. its angle of projection is tan^-1 (2). Then what is its initial velocity.

one second after projection, a stone moves at an angle 45° with horizontal. two second after projection, it moves horizontally. its angle of projection is tan^-1 (2). Then what is its initial velocity.

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one second after projection, a stone moves at an angle 45° with horizontal. two second after projection, it moves horizontally. its angle of projection is tan^-1 (2). Then what is its initial velocity.

one second after projection, a stone moves at an angle 45° with horizontal. two second after projection, it moves horizontally. its angle of projection is tan^-1 (2). Then what is its initial velocity.

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