$A$ particle is thrown with a speed of $12 \, m/s$ at an angle $60^o$ with the horizontal. The time interval between the moments when its speed is $10 \, m/s$ is $(g = 10 \, m/s^2)$.........$s$

$A$ body is projected at an angle of $45^{\circ}$ from a point on the ground at a distance of $30 \,m$ from the foot of a vertical pole of height $20 \,m$. The body just crosses the top of the pole and strikes the ground at a distance $s$ from the foot of the pole on the other side. Then, $s$ is: (in $\,m$)

$A$ body of mass $m$ thrown up vertically with velocity $v_1$ reaches a maximum height $h_1$ in $t_1$ seconds. Another body of mass $2m$ is projected with a velocity $v_2$ at an angle $\theta$. The second body reaches a maximum height $h_2$ in time $t_2$ seconds. If $t_1 = 2t_2$,then the ratio $\left(\frac{h_1}{h_2}\right)$ is

$A$ body projected with some velocity at an angle $45^{\circ}$ with the horizontal from the origin in $XY$-plane passes through a point at $(4, 3) \ m$. Its horizontal range is (in $m$)

$A$ player kicks a ball at a speed of $20 \ ms^{-1}$ so that its horizontal range is maximum. Another player $24 \ m$ away in the direction of the kick,starts running in the same direction at the same instant the ball is kicked. If he has to catch the ball just before it reaches the ground,he should run with a velocity equal to: (Take $g = 10 \ ms^{-2}$)

$A$ body is projected at an angle of $45^{\circ}$ with the horizontal with a velocity of $60 \sqrt{2} \ m/s$. Then the angle made by its velocity with the horizontal after $6 \ s$ is (in $^{\circ}$)