A projectile is thrown into space so as to have maximum horizontal range $R$. Taking the point of projection as origin, the coordinates of the points where the speed of the particle is minimum are-
$(R, R)$
$\left( {R,\frac{R}{2}} \right)$
$\left( {\frac{R}{2},\frac{R}{4}} \right)$
$\left( {R,\frac{R}{4}} \right)$
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
ball is thrown from a point with a speed $‘v_0$’ at an elevation angle of $\theta $ . From the same point and at the same instant, a person starts running with a constant speed $\frac{{'{v_0}'}}{2}$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection $\theta $ ?
A point $P$ moves in counter clock wise direction on a circular path as shown in figure. The movement of $'P'$ is such that it sweeps out a length $S = t^3 + 5$, where $'S'$ is in meter and $t$ is in seconds. The radius of the path is $20\, m$. The acceleration of $'P'$ when $t = 2\, sec$. is nearly ......... $m/s^2$
A particle is projected from a horizontal plane such that its velocity vector at time $t$ is given by $\vec v = a\hat i + (b - ct)\hat j$ . Its range on the horizontal plane is given by
A body of mass $m\, kg$ is rotating in a vertical circle at the end of a string of length $r$ metre. The difference in the kinetic energy at the top and the bottom of the circle is