$A$ particle is projected at an angle $\theta$ with the horizontal from the ground. The slope $(m)$ of the trajectory of the particle varies with time $(t)$ as:

$A$ projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3 \, ms^{-2}$ for $0.5 \, minutes$. If the maximum height reached by it is $80 \, m$,then the angle of projection is (Take $g = 10 \, ms^{-2}$)

$A$ ball having kinetic energy $KE$ is projected at an angle of $60^{\circ}$ from the horizontal. What will be the kinetic energy of the ball at the highest point of its flight?

In a sports event,a disc is thrown such that it reaches its maximum range of $80 \ m$. The distance travelled in the first $3 \ s$ is $(g = 10 \ m/s^2)$. (in $m$)

Three balls of same masses are projected with equal speeds at angles $15^{\circ}, 45^{\circ}$,and $75^{\circ}$. If their ranges are $R_1, R_2$,and $R_3$ respectively,then:

$A$ 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