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

\sqrt{(v c o s θ)^{2} + (v s i n θ)^{2}}

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\sqrt{(v c o s θ - v s i n θ)^{2} - g t}

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\sqrt{v^{2} + g^{2} t^{2} - (2 v s i n θ) g t}

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\sqrt{v^{2} + g^{2} t^{2} - (2 v c o s θ) g t}

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Solution

The correct option is C $\sqrt{v^{2} + g^{2} t^{2} - (2 v s i n θ) g t}$
Instantaneous velocity of rising mass after t sec will be
$v_{t} = \sqrt{v_{x}^{2} + v_{y}^{2}}$
Where $v_{x} = v c o s θ$ =Horizontal component of velocity
$v_{y} = v s i n θ - g t = V e r t i c a l c o m p o n e n t o f v e l o c i t y$
$v_{t} = \sqrt{(v c o s θ)^{2} + (v s i n θ - g t)^{2}}$
$v_{t} = \sqrt{v^{2} + g^{2} t^{2} - 2 v s i n θ g t}$

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Similar questions

θ

θ

v

θ

t

\to u

h

Column-I	Column-II
(a)	The speed of body at any time $t$ is	(p)	$\sqrt{\frac{2 g h}{u^{2}}}$
(b)	It will strike the ground after time $t$ equal to	(q)	$\sqrt{\frac{2 h}{g}}$
(c)	The path followed by the body is expressed as $y$ equal to	(r)	$- \frac{1}{2} g \frac{x^{2}}{u^{2}}$
(d)	The tangent of the angle $θ$ made by velocity vector striking the ground with horizontal is equal to	(s)	$\sqrt{u^{2} + g^{2} t^{2}}$

30^{\circ}

60^{\circ}

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