$A$ person throws balls into the air vertically upward at regular intervals of time of $1 \,s$. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is (Assume $g = 10 \,m/s^2$) (in $\,m$)

$A$ balloon starts rising from the ground with an acceleration of $1.25\, m/s^2$. After $8\,s$,a stone is released from the balloon. The stone will $(g = 10\, m/s^2)$:

Galileo's law of odd numbers: "The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, $1: 3: 5: 7 \ldots$]". Prove it.

$A$ 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 projected upwards. The times corresponding to height $h$ while ascending and while descending are $t_1$ and $t_2$ respectively. The velocity of projection will be

$A$ ball is dropped from a bridge of $122.5 \ m$ above a river. After the ball has been falling for $2 \ s$,a second ball is thrown straight down after it. What is the initial velocity of the second ball so that both hit the water at the same time (in $m/s$)?