$A$ particle is projected from a horizontal plane such that its velocity vector at time $t$ is given by $\vec{v} = a\hat{i} + (b - ct)\hat{j}$. Its range on the horizontal plane is given by

$A$ ball projected at an angle of $45^{\circ}$ with the horizontal crosses two points at equal heights separated by a distance at times $2 \ s$ and $8 \ s$ respectively. The horizontal distance between the two points is (Acceleration due to gravity $= 10 \ m/s^2$) (in $m$)

$A$ body is projected from the ground with a velocity $v = (3 \hat{i} + 10 \hat{j}) \text{ m/s}$. The maximum height attained and the range of the body respectively are (given $g = 10 \text{ m/s}^2$):

$A$ ball is thrown from a point with a speed $v_0$ at an elevation angle of $\theta$. From the same point and at the same instant,a person starts running with a constant speed $\frac{v_0}{2}$ to catch the ball. Will the person be able to catch the ball? If yes,what should be the angle of projection $\theta$?

If the time of flight of a projectile is $10 \ s$,and its range is $500 \ m$,then the maximum height attained by it will be ......... $m$.

$A$ stone projected from the ground with a velocity $50 \ m/s$ at an angle of $30^{\circ}$ with the horizontal crosses a wall after a time of $3 \ s$. Then the horizontal distance beyond the wall that the stone strikes the ground is (acceleration due to gravity $g = 10 \ m/s^2$) (in $m$)