At $t = 0$ a projectile is fired from a point $O$(taken as origin) on the ground with a speed of $50\,\, m/s$ at an angle of $53^o$ with the horizontal. It just passes two points $A \& B$ each at height $75 \,\,m$ above horizontal as shown The distance (in metres) of the particle from origin at $t = 2$ sec.

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., $45^{o})$ ?

Two projectiles $A$ and $B$ are thrown with the same speed such that $A$ makes angle $\theta$ with the horizontal and $B$ makes angle $\theta$ with the vertical, then

If $R$ and $H$ represent the horizontal range and the maximum height achieved by a projectile then which of the relation exists?

A projectile is projected with kinetic energy $K$. If it has the maximum possible horizontal range, then its kinetic energy at the highest point will be ......... $K$

If T is the total time of flight, $h$ is the maximum height $ \& R$ is the range for horizontal motion, the $x$ and $y$ co-ordinates of projectile motion and time $t$ are related as