Mht Cet
Motion
MHT CET 2026 19th April Evening Shift
MCQ+1 / -02026
The equation of the trajectory of a ball projected at an angle $\theta$ with the horizontal, is given as $y = x - \dfrac{gx^2}{2}$
The initial velocity of the ball is
[Given : $\tan 45^\circ = 1$, $\cos 45^\circ = \dfrac{1}{\sqrt{2}}$ ]
The initial velocity of the ball is
[Given : $\tan 45^\circ = 1$, $\cos 45^\circ = \dfrac{1}{\sqrt{2}}$ ]