Two balls $A$ and $B$ are placed at the top of 180 m tall tower. Ball $A$ is released from the top at $t = 0 \mathrm {~s}$. Ball $B$ is thrown vertically down with an initial velocity $u$ at $t = 2 \mathrm {~s}$. After a certain time, both balls meet 100 m above the ground. Find the value of $u$ in $\mathrm { m } \mathrm { s } ^ { - 1 }$. [use $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$]
(1) 10
(2) 15
(3) 20
(4) 30

From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s . A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s . A third ball released, from the rest from the same location, will reach the ground in $\_\_\_\_$ s.

A ball is projected vertically upward with an initial velocity of $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $t = 0 \mathrm {~s}$. At $t = 2 \mathrm {~s}$, another ball is projected vertically upward with same velocity. At $t =$ $\_\_\_\_$ s, second ball will meet the first ball $\left( \mathrm { g } = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 } \right)$.

Two trains $A$ and $B$ of length $l$ and $4l$ are travelling into a tunnel of length $L$ in parallel tracks from opposite directions with velocities $108 \mathrm{~km~h}^{-1}$ and $72 \mathrm{~km~h}^{-1}$, respectively. If train $A$ takes 35 s less time than train $B$ to cross the tunnel then, length $L$ of tunnel is: (Given $L = 60l$)
(1) 1200 m
(2) 900 m
(3) 1800 m
(4) 2700 m

For a train engine moving with speed of $20 \mathrm {~ms} ^ { - 1 }$, the driver must apply brakes at a distance of 500 m before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed $\sqrt { x } \mathrm {~ms} ^ { - 1 }$. The value of $x$ is $\_\_\_\_$ . (Assuming same retardation is produced by brakes)

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $( t + 1 )$ s are 125 m and $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, respectively. The distance travelled by the particle in $( t + 2 ) ^ { \text {th} } \mathrm { s }$ is $\_\_\_\_$ m.

A body moves on a frictionless plane starting from rest. If $S_n$ is distance moved between $t = n-1$ and $t = n$ and $S_{n-1}$ is distance moved between $t = n-2$ and $t = n-1$, then the ratio $\frac{S_{n-1}}{S_n}$ is $\left(1 - \frac{2}{x}\right)$ for $n = 10$. The value of $x$ is $\_\_\_\_$.