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