If the range of a gun which fires a shell with muzzle speed $V$ is $R$,then the angle of elevation of the gun is

$A$ projectile is given an initial velocity of $(3 \hat{i} + 4 \hat{j}) \text{ m s}^{-1}$, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. Assuming $g = 10 \text{ m s}^{-2}$, if the equation of its trajectory can be written as $\frac{1}{9} [\beta x + \gamma x^2]$, then the value of $\gamma$ 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:

$A$ ball is thrown from the location $(x_0, y_0) = (0, 0)$ of a horizontal playground with an initial speed $v_0$ at an angle $\theta_0$ from the $+x$-direction. The ball is to be hit by a stone,which is thrown at the same time from the location $(x_1, y_1) = (L, 0)$. The stone is thrown at an angle $(180^{\circ} - \theta_1)$ from the $+x$-direction with a suitable initial speed $v$. For a fixed $v_0$,when $(\theta_0, \theta_1) = (45^{\circ}, 45^{\circ})$,the stone hits the ball after time $T_1$,and when $(\theta_0, \theta_1) = (60^{\circ}, 30^{\circ})$,it hits the ball after time $T_2$. In such a case,$(T_1 / T_2)^2$ is. . . . .

Two seconds after projection,a projectile is travelling in a direction inclined at $30^o$ to the horizontal. After one more second,it is travelling horizontally. What is the magnitude and direction of its velocity at the initial point?

$A$ particle is projected at an angle of $60^{\circ}$ with the horizontal from the ground with a velocity $10 \sqrt{3} \ m/s$. The angle between the velocity vector after $2 \ s$ and the initial velocity vector is $(g = 10 \ m/s^2)$. (in $^{\circ}$)