The rate of change of the volume of a sphere of radius $r$ with respect to its surface area is:

$A$ stone thrown vertically upwards rises $s$ metres in $t$ seconds,where $s = 80t - 16t^2$. Then,the velocity after $2$ seconds is .......... $m/s$.

$A$ kite is $120 \ m$ high and $130 \ m$ of string is out. If the kite is moving away horizontally at the rate of $39 \ m/sec$,then the rate at which the string is being paid out is:

If displacement $s = 5 \sin(2t)$,then the velocity at the end of $t = \frac{\pi}{3} \text{ s}$ is

$A$ stone is dropped in a quiet lake and it is observed that waves move in circles. If the radius of a circular wave increases at the rate of $2 \text{ cm/sec}$, then the rate of increase in its area at the instant when its radius is $10 \text{ cm}$, is in $\text{cm}^2\text{/sec}$: (in $\pi$)

$A$ ball thrown vertically upwards falls back on the ground after $6 \, s$. Assuming that the equation of motion is of the form $s = ut - 4.9t^2$,where $s$ is in metres and $t$ is in seconds,find the initial velocity $u$ at $t = 0$ in $m/s$.