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(C) The initial velocity $v$ is resolved into two components: Horizontal component: $v_x = v\cos	heta$ Vertical component: $v_y = v\sin	heta$ After

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$\sqrt{v^2 + g^2t^2 - 2vgt\sin	heta}$

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(C) The initial velocity $v$ is resolved into two components: Horizontal component: $v_x = v\cos	heta$ Vertical component: $v_y = v\sin	heta$ After $t$ seconds,the horizontal component remains constant: $v_x = v\cos	heta$ The vertical component changes due to gravity: $v_y = v\sin	heta - gt$ The magnitude of the resultant velocity $v_t$ is given by: $v_t = \sqrt{v_x^2 + v_y^2}$ $v_t = \sqrt{(v\cos	heta)^2 + (v\sin	heta - gt)^2}$ $v_t = \sqrt{v^2\cos^2	heta + v^2\sin^2	heta + g^2t^2 - 2vgt\sin	heta}$ Since $\cos^2	heta + \sin^2	heta = 1$,we get: $v_t = \sqrt{v^2 + g^2t^2 - 2vgt\sin	heta}$

$A$ body of mass $m$ is thrown upwards at an angle $\theta$ with the horizontal with velocity $v$. While rising up,the velocity of the mass after $t$ seconds will be:

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