$A$ stone is thrown vertically upward with an initial velocity $v_0$. The distance travelled in time $t = \frac{1.5 v_0}{g}$ 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 $m$)

$A$ particle is dropped from the top of a tower. The distance covered by it in the last one second is equal to the distance covered by it in the first three seconds. The height of the tower is $....m$

$A$ ball is released from a height $h$. If $t_{1}$ and $t_{2}$ are the times required to complete the first half and the second half of the distance respectively,then choose the correct relation between $t_{1}$ and $t_{2}$.

$A$ stone dropped from the top of the tower touches the ground in $4 \, s$. The height of the tower is about..........$m$.

$A$ stone is thrown upwards with a velocity $50 \,ms^{-1}$. Another stone is simultaneously thrown downwards from the same location with a velocity $50 \,ms^{-1}$. When the first stone is at the highest point, the velocity of the second stone is $...... \,ms^{-1}$ (Take $g = 10 \,ms^{-2}$).