From a balloon rising vertically with a uniform velocity of $v \ ft/sec$,a stone is dropped. If the stone reaches the ground after $4 \ sec$,what is the height of the balloon above the ground at that moment (in $ft$)? (Take $g = 32 \ ft/sec^2$)

The rate of change of the surface area of a sphere of radius $r$ when the radius is increasing at the rate of $2 \text{ cm/sec}$ is proportional to

$A$ stone,thrown vertically upward from the surface of the moon at a velocity of $24 \ m/s$,reaches a height of $s = 24t - 0.8t^2$ meters after $t$ seconds. The acceleration due to gravity in $m/s^2$ at the surface of the moon is:

$A$ ladder $20 \ ft$ long leans against a vertical wall. The top end slides downwards at the rate of $2 \ ft/sec$. The rate at which the lower end moves on a horizontal floor when it is $12 \ ft$ from the wall is

$A$ stone is dropped into a quiet lake and waves move in circles at a speed of $4 \text{ cm/s}$. At the instant when the radius of the circular wave is $10 \text{ cm}$,how fast is the enclosed area increasing?

By dropping a stone in a quiet lake,a wave in the form of a circle is generated. The radius of the circular wave increases at the rate of $2.1 \text{ cm/sec}$. Then the rate of increase of the enclosed circular region,when the radius of the circular wave is $10 \text{ cm}$,is (Given $\pi = \frac{22}{7}$)