$A$ ball is thrown at an angle $\theta$ with the horizontal. Its horizontal range is equal to its maximum height. This is possible only when the value of $\tan \theta$ is ..........

$A$ projectile is launched from the ground, such that it hits a target on the ground which is $90 \,m$ away. The minimum velocity of the projectile to hit the target is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,ms^{-1}$)

$A$ projectile fired with initial velocity $u$ at an angle $\theta$ has a range $R$. If the initial velocity is doubled at the same angle of projection,then the new range will be:

$A$ body is projected at an angle of $45^{\circ}$ with the horizontal with a velocity of $60 \sqrt{2} \ m/s$. Then the angle made by its velocity with the horizontal after $6 \ s$ is (in $^{\circ}$)

The velocity of projection of a body is increased by $2 \%$. Other factors remaining unchanged,what will be the percentage change in the maximum height attained? (in $\%$)

$A$ ball $P$ is dropped vertically and another ball $Q$ is thrown horizontally with the same initial vertical velocity (which is $0$) from the same height $h$ at the same time. If air resistance is neglected,then: