A projectile is launched at an angle ' $\alpha$ ' with the horizontal with a velocity $20 \; ms ^{-1}$. After $10 s$, its inclination with horizontal is ' $\beta$ '. The value of $\tan \beta$ will be : $\left( g =10 \; ms ^{-2}\right)$

The equation of motion of a projectile are given by $ x = 36 t\,metre$ and $2y = 96 t -9.8 t^2\, metre$. The angle of projection is

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., $45^{o})$ ?

Define projectile motion and projectile particle.