The horizontal range is four times the maximum height attained by a projectile. The angle of projection is  .......... $^o$

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.

Given below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason $R$.

Assertion A :Two identical balls $A$ and $B$ thrown with same velocity '$u$ ' at two different angles with horizontal attained the same range $R$. If $A$ and $B$ reached the maximum height $h_{1}$ and $h_{2}$ respectively, then $R =4 \sqrt{ h _{1} h _{2}}$

Reason R: Product of said heights.

$h _{1} h _{2}=\left(\frac{u^{2} \sin ^{2} \theta}{2 g }\right) \cdot\left(\frac{u^{2} \cos ^{2} \theta}{2 g }\right)$

Choose the $CORRECT$ answer 

Two projectiles are thrown simultaneously in the same plane from the same point. If their velocities are $v_1$ and $v_2$ at angles $\theta _1$ and $\theta_2$ respectively from the horizontal, then answer the following question

If $v_1\,\,sin\,\,\theta _1 = v_2\,\,sin\,\,\theta _2$, then choose the incorrect statement

Two balls are projected from the same point simultaneously.First ball is projected vertically upwards and the second ball at an angle of projection $60^o$ to the ground level. Both the balls reach the ground simultaneously. The ratio of their velocities are

The velocity at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times its initial velocity of projection $(u)$. Its range on the horizontal plane is .............