$A$ ball is thrown vertically upwards with an initial velocity $u$ and reaches its maximum height in $5 \,s$. The ratio of the distance travelled by the ball in the $2^{nd}$ and $7^{th}$ second is (assume $g=10 \,m/s^2$):

$A$ body projected vertically upwards crosses a point twice in its journey at a height $h$ after $t_1$ and $t_2$ seconds. The maximum height reached by the body is

$A$ body falling from a vertical height of $10 \, m$ pierces through a distance of $1 \, m$ in sand. It faces an average retardation in sand equal to (where $g$ = acceleration due to gravity).

$A$ ball is projected upwards from a height $h$ above the surface of the earth with velocity $v$. The time at which the ball strikes the ground is

Two bodies of different masses $m_a$ and $m_b$ are dropped from two different heights $a$ and $b$. The ratio of the time taken by the two to cover these distances is

$A$ ball is projected vertically up from the ground. Boy $A$, standing at the window of the first floor of a nearby building, observes that the time interval between the ball crossing him while going up and the ball crossing him while going down is $2 \ s$. Another boy $B$, standing on the second floor, notices the time interval between the ball passing him twice, during up and down motion, is $1 \ s$. Calculate the difference between the vertical positions of boy $B$ and boy $A$. (Assume $g = 10 \ m \ s^{-2}$) (in $m$)