$A$ body is projected at an angle of $45^{\circ}$ with the horizontal with a velocity of $60 \sqrt{2} \ m/s$. Then the angle made by its velocity with the horizontal after $6 \ s$ is (in $^{\circ}$)

$A$ projectile has a range of $1.5 \ km$ when projected at an angle of $15^{\circ}$. Find the range of the projectile when it is projected at an angle of $45^{\circ}$ with the same velocity. (in $km$)

If the equation of motion of a projectile is $y=Ax-Bx^2$,then the ratio of the maximum height reached and the range of the projectile is

$A$ body is projected with a speed $u \ m/s$ at an angle $\beta$ with the horizontal. The kinetic energy at the highest point is $3/4$ of the initial kinetic energy. The value of $\beta$ is: (in $^{\circ}$)

$A$ particle is projected at $60^\circ$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is

The equation for the trajectory of a projectile is $y = \left( \frac{x}{\sqrt{3}} - \frac{x^2}{60} \right) \text{ m}$. The velocity of projection of the projectile is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$) (in $\text{ m s}^{-1}$)