$A$ body thrown vertically up with initial velocity $52 \,m/s$ from the ground passes twice a point at $h$ height above at an interval of $10 \,s$. The height $h$ is .........$m$ $(g=10 \,m/s^2)$.

Time taken by an object falling from rest to cover the height of $h_1$ and $h_2$ is respectively $t_1$ and $t_2$,then the ratio of $t_1$ to $t_2$ is:

$A$ ball is thrown vertically upwards with a velocity of $15 \ m/s$. To what maximum height will it reach? (Take $g = 10 \ m/s^2$)

$A$ 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 $m$)

$A$ 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

$A$ body thrown vertically upwards from the ground reaches a maximum height $H$. The ratio of the velocities of the body at heights $\frac{3H}{4}$ and $\frac{8H}{9}$ from the ground is