Shots are fired from the top of a tower and from its bottom simultaneously at angles $30^o$ and $60^o$ as shown. If horizontal distance of the point of collision is at a distance $'a'$ from the tower then height of tower $h$ is :

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is $v$, the total area around the fountain that gets wet is :

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.

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

A projectile is fired at an angle of $30^{\circ}$ to the horizontal such that the vertical component of its initial velocity is $80\,m / s$. Its time of flight is $T$. Its velocity at $t=\frac{T}{4}$ has a magnitude of nearly $........\frac{m}{s}$

For a given velocity, a projectile has the same range $R$ for two angles of projection if $t_1$ and $t_2$ are the times of flight in the two cases then