Q. 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 $t$ second will be
A) $\sqrt{\left( v\cos\theta \right)^{2} + \left( v\sin\theta \right)^{2}}$
B) $\sqrt{\left( v\cos\theta\ - v\sin\theta \right)^{2} - gt}$
C) $\sqrt{v^{2} + g^{2}t^{2} - \left( 2\ v\sin\theta \right)gt}$
D) $\sqrt{v^{2} + g^{2} + g^{2} - \left( 2\ v\cos\theta \right)gt}$
Explanation
(c) Instantaneous velocity of running mass after $t$ sec will be
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v}_{t} = \sqrt{v_{x}^{2} + v_{y}^{2}}$
Where, $v_{x} = v\sin\theta - gt =$ vertical component of velocity,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{y} = v\cos\theta =$horizontal component of velocity
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{t} = \sqrt{\left( v\cos\theta \right)^{2} + \left( v\sin{\theta - gt} \right)^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{t} = \sqrt{v^{2} + g^{2}t^{2} - 2v\sin\theta gt}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.