If the instantaneous velocity of a particle projected as shown in the figure is given by $v = a \hat{i} + (b - ct) \hat{j}$,where $a, b$,and $c$ are positive constants,the range on the horizontal plane will be

If the velocity of a projectile at its maximum height is $\frac{1}{\sqrt{2}}$ times its initial velocity,what is its range?

$A$ body is projected at an angle $\theta$ so that its range is maximum. If $T$ is the time of flight,then the value of maximum range is (acceleration due to gravity $= g$)

$A$ body is projected at an angle of $60^{\circ}$ with the horizontal such that the vertical component of its initial velocity is $40 \ m \ s^{-1}$. The magnitude of velocity of the projectile at one quarter of its time of flight is nearly (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m \ s^{-1}$)

$A$ 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. . . . .

$A$ football player throws a ball with a velocity of $50 \ m/s$ at an angle of $30^{\circ}$ from the horizontal. The ball remains in the air for ...... $s$ $(g = 10 \ m/s^2)$.