$A$ particle is projected from point $O$ with velocity $u$ at an angle $\alpha$ with the horizontal. If at point $P$,its velocity is perpendicular to the initial direction of projection,then find the time taken to reach from $O$ to $P$.

Two balls $A$ and $B$ are projected at an angle of $45^{\circ}$ and $60^{\circ}$ respectively, so that the maximum heights reached are same for both. The ratio of initial velocity of projection of ball $A$ to that for ball $B$ is $(\sin 30^{\circ} = \cos 60^{\circ} = 1/2, \sin 45^{\circ} = \cos 45^{\circ} = 1/\sqrt{2}, \sin 60^{\circ} = \cos 30^{\circ} = \sqrt{3}/2)$.

The equation of a projectile is $y = \sqrt{3}x - \frac{x^2}{2}$. The velocity of projection is:

$A$ body is projected at an angle of $60^{\circ}$ with the horizontal such that the vertical component of its initial velocity is $40 \ m \ s^{-1}$. The magnitude of velocity of the projectile at one quarter of its time of flight is nearly (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m \ s^{-1}$)

If the initial velocity of a projectile is doubled,keeping the angle of projection the same,the maximum height reached by it will:

At the top of the trajectory of a projectile,the acceleration is