$A$ 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 $km$)

$A$ projectile is thrown with an initial velocity of $(a \hat{i} + b \hat{j}) \text{ m s}^{-1}$. If the range of the projectile is twice the maximum height reached by it,then

If the time of flight of a bullet over a horizontal range $R$ is $T \ s$,the inclination of the direction of projection to the horizontal is

$A$ body of mass $10 \ kg$ is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20, 10)$. If $T$ is the time of flight,then its momentum vector,at time $t = \frac{T}{\sqrt{2}}$,is. [Take $g = 10 \ m/s^2$]

$A$ projectile is given an initial velocity of $(\hat{i} + 2\hat{j}) \, m/s$,where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g = 10 \, m/s^2$,the equation of its trajectory is:

For a given velocity,a projectile has the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then: