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 $ ?

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)

During which time interval is the particle described by these position graphs at rest?

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)

The co-ordinates of a moving particle at a time $t$, are give by, $x = 5 sin 10 t, y = 5 cos 10t$. The speed of the particle is :

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