$A$ ball is projected from the ground with a speed $15 \, m/s$ at an angle $\theta$ with the horizontal such that its range and maximum height are equal. Then,$\tan \theta$ will be equal to:

For a projectile motion, the range $R$ is '$n$' times the maximum height $H$. So the angle of projection is

The horizontal range of a projectile projected at an angle of $45^{\circ}$ with the horizontal is $50 \ m$. The height of the projectile when its horizontal displacement is $20 \ m$ is (in $m$)

$A$ projectile is fired with a speed $u$ at an angle $\theta$ with the horizontal. Its speed when its direction of motion makes an angle $\alpha$ with the horizontal is

$A$ body is projected horizontally from the top of a tower with an initial velocity of $18 \, m s^{-1}$. It hits the ground at an angle of $45^{\circ}$. What is the vertical component of velocity when it strikes the ground? $(ms^{-1})$

The relation between the horizontal displacement $x$ (in metre) and the vertical displacement $y$ (in metre) of a projectile is $y = 3x - 0.8x^2$. The time of flight of the projectile is (Acceleration due to gravity $g = 10 \ m/s^2$). (in $s$)