The range of a particle when launched at an angle of $15^\circ$ with the horizontal is $1.5 \, km$. What is the range of the projectile when launched at an angle of $45^\circ$ to the horizontal? (in $km$)

$A$ body is projected at $t = 0$ with a velocity $10\,ms^{-1}$ at an angle of $60^\circ$ with the horizontal. The radius of curvature of its trajectory at $t = 1\,s$ is $R$. Neglecting air resistance and taking acceleration due to gravity $g = 10\,ms^{-2}$,the value of $R$ is ........ $m$.

$A$ projectile is given an initial velocity of $(\hat i + 2\hat j) \ ms^{-1}$,where $\hat i$ is along the ground and $\hat j$ is along the vertical. If $g = 10 \ m/s^2$,the equation of its trajectory is

$A$ particle is thrown with a velocity of $u \, m/s$ at an angle of $30^{\circ}$ with the horizontal. It passes through points $A$ and $B$ at the same vertical height at times $t_1 = 1 \, s$ and $t_2 = 3 \, s$ respectively,as shown in the figure. The value of $u$ is ....... $m/s$ (take $g = 10 \, m/s^2$).

$A$ projectile is projected with a velocity of $25 \, m/s$ at an angle $\theta$ with the horizontal. After $t$ seconds,its inclination with the horizontal becomes zero. If $R$ represents the horizontal range of the projectile,the value of $\theta$ will be: [use $g = 10 \, m/s^2$]

$A$ ball is projected with a velocity of $10 \ ms^{-1}$ at an angle of $60^{\circ}$ with the vertical direction. Its speed at the highest point of its trajectory will be $............... \ ms^{-1}$.