The horizontal range is four times the maximum height attained by a projectile. The angle of projection is ......... $^o$

A cricket fielder can throw the cricket ball with a speed $v_{0} .$ If he throws the ball while running with speed $u$ at an angle $\theta$ to the horizontal, find

$(a)$ the effective angle to the horizontal at which the ball is projected in air as seen by a spectator

$(b)$ what will be time of flight?

$(c)$ what is the distance (horizontal range) from the point of projection at which the ball will land ?

$(d)$ find $\theta$ at which he should throw the ball that would maximise the horizontal range as  found in $(iii)$.

$(e)$ how does $\theta $ for maximum range change if $u > u_0$. $u =u_0$ $u < v_0$ ?

$(f)$ how does $\theta $ in $(v)$ compare with that for $u=0$ $($ i.e., $45^{o})$ ?

A body is projected at an angle $\theta $ with horizontal, another body is projected with the same speed at an angle $\theta $ with the vertical then the ratio of the maximum height is

Two bodies are thrown up at angles of $45^o $ and $60^o $, respectively, with the horizontal. If both bodies attain same vertical height, then the ratio of velocities with which these are thrown is

A particle projected from ground moves at angle $45^{\circ}$ with horizontal one second after projection and speed is minimum two seconds after the projection. The angle of projection of particle is [Neglect the effect of air resistance]

A particle $A$ is projected vertically upwards. Another identical particle $B$ is projected at an angle of $45^o $ . Both reach the same height. The ratio of the initial kinetic energy of $A$ to that of $B$ is