A body of mass m is thrown upwards at an angle q  with the…

Physics Motion in a Plane Time For Max Height Single Correct MCQ

Published on: September 12, 2026

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

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

\(\sqrt{(v \cos \theta - v_0 \sin \theta)^2 - gt}\)

\(\sqrt{v^2} = g^2 t^2 - (2 v g n \sigma) g t\)

\(\sqrt{v^2 + g^2 t^2 - 2 v \cos \theta \, g t}\)

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The correct answer is:

Instantaneous velocity of rising mass after t sec will be \(v_o = \sqrt{v_x^2 + v_y^2}\)

where \(v_s \quad v \cos \psi\) Horizontal component of velocity

\(v_y \quad v \sin \theta - gt\) Vertical component of velocity

\(v_c = \sqrt{(v \cos \theta)^2 + (v \sin \theta)^2 - g a^2}\)

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

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