$A$ ball is dropped from a high-rise platform at $t=0$ starting from rest. After $6$ seconds,another ball is thrown downwards from the same platform with a speed $v$. The two balls meet at $t=18\,s$. What is the value of $v$ in $m/s$? (Take $g= 10\,m/s^2$)

Suggest a suitable physical situation for the following $v-t$ graph.

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

$A$ body is thrown vertically upwards. If air resistance is to be taken into account,then the time during which the body rises is

$A$ ball is thrown vertically upwards with a velocity of $20 \; m s^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25.0 \; m$ from the ground. How high will the ball rise above the point of projection (in $; m$)? Take $g = 10 \; m s^{-2}$.

$A$ body is thrown vertically upwards and takes $5 \text{ s}$ to reach maximum height. The distance travelled by the body will be same in