The vector sum of two forces is perpendicular to their vector differences. In that case, the forces

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 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 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-

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 

A particle is moving on a circular path of radius $r$ with uniform velocity $v$. The change in velocity when the particle moves from $P$ to $Q$ is $(\angle POQ = {40^o})$