The range of a particle when launched at an angle of $15^\circ$ with the horizontal is $1.5 \, km$. What is the range of the projectile when launched at an angle of $45^\circ$ to the horizontal? (in $km$)

The equation of trajectory of a projectile is $y = 10x - (5/9)x^2$. If we assume $g = 10 \ m/s^2$, the range of the projectile (in meters) is:

$A$ body is projected from the ground at an angle of $45^{\circ}$ with the horizontal. Its velocity after $2 \, s$ is $20 \, m/s$. The maximum height reached by the body during its motion is $m$. (use $g = 10 \, m/s^2$)

An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$, where $h$ is the height, $x$ is the horizontal distance, and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

$A$ projectile crosses two walls of equal height $H$ symmetrically as shown. The maximum height of the projectile is ........ $m$.

Two projectiles of same mass and with same velocity are thrown at an angle $60^o$ and $30^o$ with the horizontal,then which quantity will remain same?