A bullet is fired from a cannon with velocity $500 \,m/s$. If the angle of projection is ${15^o}$ and $g = 10m/{s^2}$. Then the range is

A stone projected with a velocity u at an angle $\theta$ with the horizontal reaches maximum height $H_1$. When it is projected with velocity u at an angle $\left( {\frac{\pi }{2} - \theta } \right)$ with the horizontal, it reaches maximum height $ H_2$. The relation between the horizontal range R of the projectile, $H_1$ and $H_2$ is

If a stone is to hit at a point which is at a distance $d$ away and at a height $h$ above the point from where the stone starts, then what is the value of initial speed $u$ if the stone is launched at an angle $\theta $ ?

A particle is projected from a horizontal plane ($x-z$ plane) such that its velocity vector at time t is given by $\vec V = a\hat i + (b - ct)\hat j$ Its range on the horizontal plane is given by

A bullet fired at an angle of $30^o$ with the horizontal hits the ground $3.0\; km$ away. By adjusting its angle of projection, can one hope to hit a target $5.0\; km$ away ? Assume the muzzle speed to be fixed, and neglect air resistance.

The range of the projectile projected at an angle of $15^{\circ}$ with horizontal is $50\,m$. If the projectile is projected with same velocity at an angle of $45^{\circ}$ with horizontal, then its range will be $........\,m$