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

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$\sqrt{v^{2} + g^{2} t^{2} - (2 v \sin θ) g t}$

$\sqrt{v^{2} + g^{2} t^{2} - (2 v \cos θ) g t}$

$\sqrt{(v \cos θ)^{2} + (v \sin θ)^{2}}$

$\sqrt{(v \cos θ - v \sin θ)^{2} - g t}$

answer is C.

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Detailed Solution

Instantaneous velocity of rising mass after t second will be

$v_{t} = \sqrt{{v_{x}}^{2} + {v_{y}}^{2}}$

where, v_x = v cos $θ$ = horizontal component of velocity
v_y = v sin $θ$ $-$ gt = vertical component of velocity.

$\Rightarrow v_{t} = \sqrt{(v \cos θ)^{2} + (v \sin θ - g t)^{2}}$

$\Rightarrow v_{t} = \sqrt{v^{2} + g^{2} t^{2} - (2 v \sin θ) g t}$

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