The position of a projectile launched from the origin at $t = 0$ is given by $\vec{r} = (40\hat{i} + 50\hat{j})\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10\,m/s^2$)

$A$ particle is projected with speed $u$ at an angle $\theta$ with the horizontal from the ground. If it is at the same height from the ground at times $t_1$ and $t_2$,then its average velocity in the time interval $t_1$ to $t_2$ is .........

$A$ projectile can have 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 their product is:

$A$ ball $A$ is projected vertically upwards with a certain initial speed $u$. Another ball $B$ of the same mass is projected at an angle of $30^{\circ}$ with the vertical with the same initial speed $u$. At the highest point,the ratio of the potential energy of ball $A$ to that of ball $B$ will be: $(\sin 90^{\circ}=1, \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2})$

If the initial velocity in the horizontal direction of a projectile is the unit vector $\hat{i}$ and the equation of the trajectory is $y = 5x(1 - x)$,find the $y$-component vector of the initial velocity. (Take $g = 10\,m/s^2$) (in $,\hat{j}$)

For a given angle of the projectile,if the initial velocity is doubled,the range of the projectile becomes: