Question

Two balloons are simultaneously released from two buildings $A$ and $B$. Balloon from $A$ rises with constant velocity of $10m{s}^{-1}$, While the other one rises with constant velocity of $20m{s}^{-1}$. Due to wind the balloons gather horizontal velocity $v_x=0.5y$, where $y$ is the height from the point of release. The buildings are at a distance of $250m$ and after some time $t$ the balloons collide. Find the difference in height (in $m$ ) of the buildings.

Answer 1

Answer: 100

$20\times t=10t+a$
$\Rightarrow 10t=a$
For $B:$
$v_x=\frac{y}{2}$
$\frac{dx}{dt}=\frac{20t}{2}$
$\underset{0}{\overset{b}{\int }}dx=\underset{0}{\overset{t}{\int }}10tdt$
$b=5t^2$.....(i)
For $A:$
$v_x=\frac{y}{2}$
$\frac{dx}{dt}=\frac{10t}{2}$
$\underset{250}{\overset{b}{\int }}dx=\underset{0}{\overset{t}{\int }}5tdt$
$b-250=\frac{5t^2}{2}$ ....(ii)
Using (i) and (ii), we get
$5t^2-250=\frac{5t^2}{2}$
$\Rightarrow t=10s$ and $a=100m$