A projectile is fired at $30^{\circ}$ to the horizontal, The vertical component of its velocity is $80 \;ms ^{-1}$, Its time flight is $T$. What will be the velocity of projectile at $t =\frac{ T }{2}$?

If time of flight of a projectile is $10$ seconds. Range is $500$ meters. The maximum height attained by it will be  ......... $m$

A ball thrown by one player reaches the other in $2\, sec$. The maximum height attained by the ball above the point of projection will be about .......... $m$

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 $g=10\,m s ^{-2}$ )

Show that for a projectile the angle between the velocity and the $x$ -axis as a function of time is given by

$\theta(t)=\tan ^{-1}\left(\frac{v_{0 y}-g t}{v_{0 x}}\right)$

Show that the projection angle $\theta_{0}$ for a projectile launched from the origin is given by

$\theta_{0}=\tan ^{-1}\left(\frac{4 h_{m}}{R}\right)$

Where the symbols have their usual meaning.

The equation of projectile is $y = 16x\, - \,\frac{{5{x^2}}}{4}$, The horizontal range is .......... $m$