$A$ projectile is thrown with a velocity of $50 \, ms^{-1}$ at an angle of $53^o$ with the horizontal. The equation of the trajectory is given by:

$A$ projectile with speed $50 \ m \ s^{-1}$ is thrown at an angle of $60^{\circ}$ with the horizontal. The maximum height that can be reached is,(acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m$)

For angles of projection of a projectile at angles $(45^\circ + \theta)$ and $(45^\circ - \theta)$,the horizontal range described by the projectile are in the ratio of:

The trajectory of a projectile near the surface of the earth is given as $y = 2x - 9x^2$. If it were launched at an angle $\theta_0$ with speed $v_0$,then $(g = 10 \, ms^{-2})$:

$A$ particle projected from the ground moves at an angle of $45^{\circ}$ with the horizontal one second after projection,and its speed is minimum two seconds after the projection. What is the angle of projection of the particle? [Neglect the effect of air resistance]

$A$ ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal from a distance $d_1$ from the foot of a pole. It just touches the top of the pole and then falls on the ground at a distance $d_2$ from the pole on the other side. The height of the pole is ...........