Two seconds after projection a projectile is travelling in a direction inclined at $30^o$ to horizontal, after one more second it is travelling horizontally. What is the magnitude and direction of its velocity at initial point
$20\sqrt {3\,} \,\,m/s\,,\,\,{30^o}$
$20\sqrt {3\,} \,\,m/s\,,\,\,{60^o}$
$10\sqrt {3\,} \,\,m/s\,,\,\,{30^o}$
$10\sqrt {3\,} \,\,m/s\,,\,\,{60^o}$
If the instantaneous velocity of a particle projected as shown in figure is given by $v =a \hat{ i }+(b-c t) \hat{ j }$, where $a, b$, and $c$ are positive constants, the range on the horizontal plane will be
Average velocity of a particle is projectile motion between its starting point and the highest point of its trajectory is : (projection speed = $u$, angle of projection from horizontal= $\theta$)
A particle is projected with a velocity $v$ such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where $g$ is acceleration due to gravity)
For a particle in uniform circular motion, the acceleration $\overrightarrow{ a }$ at any point $P ( R , \theta)$ on the circular path of radius $R$ is (when $\theta$ is measured from the positive $x\,-$axis and $v$ is uniform speed)
A body of mass $m$ is suspended from a string of length $l$. What is minimum horizontal velocity that should be given to the body in its lowest position so that it may complete one full revolution in the vertical plane with the point of suspension as the centre of the circle