A ball is dropped from rest at a height $h$ from the ground. If it travels a distance of $\frac{9h}{25}$  in its last second of motion, then value of $h$ is $\left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

Graph of  $\left(\frac{1}{V}\right)$ versus $x$ for a particle under motion is shown in figure, where $v$ is velocity and $x$ is position. Find the time taken by particle to move from $x=4 \mathrm{m}$ to $x=12 \mathrm{m}$.

The motion of a particle moving along $x$-axis is represented by the equation $\frac{dv}{dt}=6-3v$, where$v$ is in $\mathrm{m} {\mathrm{s}}^{-1}$ and$t$ is in second. If the particle is at rest at$t=0$, then

The position-time graph for two particles $\mathrm{P}$ and $\mathrm{Q}$ moving on $\mathrm{x}$-axis is shown in the figure. Then