$A$ projectile is thrown with velocity $v$ making an angle $\theta$ with the vertical. It reaches a maximum height $H$. The time of flight is:

In dealing with the motion of a projectile in the air,we ignore the effect of air resistance on the motion. This gives a trajectory that is a parabola,as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

An object of mass $m$ is projected with a momentum $p$ at such an angle that its maximum height is $\frac{1}{4}$th of its horizontal range. Its minimum kinetic energy in its path will be

$A$ stone projected with a velocity $u$ at an angle $\theta$ with the horizontal reaches maximum height $H_1$. When it is projected with velocity $u$ at an angle $(\frac{\pi}{2} - \theta)$ with the horizontal,it reaches maximum height $H_2$. The relation between the horizontal range $R$ of the projectile,$H_1$ and $H_2$ is

$A$ particle of mass $m$ is projected at $45^o$ at $V_0$ speed from point $P$ at $t = 0$. The angular momentum of the particle about $P$ at $t = \frac{V_0}{g}$ is:

$A$ stone is projected with a velocity $20 \sqrt{2} \, m/s$ at an angle of $45^{\circ}$ to the horizontal. The average velocity of the stone during its motion from the starting point to its maximum height is $.......... \, m/s$ (take $g = 10 \, m/s^2$).