$A$ particle is projected with velocity $u$ at an angle $\theta$ with the horizontal. What is the change in its velocity at the maximum height?

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

If bullets are fired in all possible directions from the same point with an equal velocity of $10 \,m \,s^{-1}$ and an angle of projection $45^{\circ}$,then the area covered by the bullets on the ground is nearly (Acceleration due to gravity $g = 10 \,m \,s^{-2}$) (in $\,m^2$)

$A$ stone is projected at an angle of $30^{\circ}$ to the horizontal. The ratio of the kinetic energy of the stone at the point of projection to its kinetic energy at the highest point of its flight will be:

$A$ 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. . . . .

$A$ projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $147 \ m/s$. Then the time after which its inclination with the horizontal is $45^{\circ}$ is ......... $s$.