$A$ ball is thrown vertically upwards from the ground $G$ with a speed $u$. It reaches a point $B$ at a height $h$ (lower than the maximum height) after time $t_1$. It returns to the ground after time $t_2$ from the instant it was at $B$ during the upward journey. Then $t_1t_2$ is equal to

$A$ ball of mass $m_1$ and another ball of mass $m_2$ are dropped from equal height. If the time taken by the balls are $t_1$ and $t_2$ respectively,then:

$A$ tennis ball is released from a height $h$ and after freely falling on a wooden floor,it rebounds and reaches a height $\frac{h}{2}$. The velocity versus height of the ball during its motion may be represented graphically by (graphs are drawn schematically and not to scale):

$A$ body is thrown vertically upwards and reaches its maximum height in $t$ seconds. The total time from the time of projection to reach a point at half of its maximum height while returning (in seconds) is:

Water drops are falling from a nozzle of a shower onto the floor,from a height of $9.8\, m$. The drops fall at a regular interval of time. When the first drop strikes the floor,at that instant,the third drop begins to fall. Locate the position of the second drop from the floor when the first drop strikes the floor. (In $m$)

$A$ ball is dropped from rest at time $t=0$ from a certain height. $A$ second ball is dropped from the same height at time $t=1 \,s$. At what time $t$, does the distance between the two balls become $10 \,m$ (in $s$)?