A projectile fired at $30^{\circ}$ to the ground is observed to be at same height at time $3 s$ and $5 s$ after projection, during its flight. The speed of projection of the projectile is $.........\,ms ^{-1}$(Given $g=10\,m s ^{-2}$ )

A boy playing on the roof of a $10\, m$ high building throws a ball with a speed of $10\, m/s$ at an angle $30^o$ with the horizontal. ........ $m$ far from the throwing point will the ball be at the height of $10\, m$ from the ground . $(g \,= \,10 m/s^2, \,sin \,30^o \,= \,\frac{1}{2}$, $\cos \,{30^o}\, = \,\frac{{\sqrt 3 }}{2}$)

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 particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution. If this particle were projected with the same speed at an angle ' $\theta$ ' to the horizontal, the maximum height attained by it equals $4 \mathrm{R}$. The angle of projection, $\theta$, is then given by

An object is thrown along a direction inclined at an angle of ${45^o}$ with the horizontal direction. The horizontal range of the particle is equal to