$A$ bullet is fired with a speed of $1000 \, m/s$ in order to hit a target $100 \, m$ away. If $g = 10 \, m/s^2$,the gun should be aimed:

$A$ projectile thrown with a speed $v$ at an angle $\theta$ has a range $R$ on the surface of the Earth. For the same $v$ and $\theta$,its range on the surface of the Moon will be:

$A$ 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 $m$)

$A$ 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 $g = 10 \, m/s^2$]

$A$ projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$,the range of the projectile is $d$. If at the highest point in its trajectory,the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$,then the new range is $d^{\prime}=n d$. The value of $n$ is. . . . .

$A$ particle reaches its highest point when it has covered exactly one half of its horizontal range. The corresponding point on the displacement-time graph is characterized by