Question

A particle is dropped from a $200m$ tall tower. At the same time, a second particle is thrown upwards with velocity $50m/s.$ Calculate when and where would these two particles meet?

Answer 1

Suppose both the bodies $y$ meet at point $C$ now the distance $h$ travelled by the body $A$ which is dropped freely from the top of the tower , using relation,
$s=ut+\frac{1}{2}a{t}^2$
$h=0+\frac{1}{2}a{t}^2$
or $h=\frac{1}{2}g{t}^2..............\left(1\right)$
Similarly , the distance covered by the body $B$,
$-(200-h)=50t+\frac{1}{2}g{t}^2$
or $-(200-h)+50t=\frac{1}{2}g{t}^2.......\left(2\right)$
Now on comparing equation $\left(1\right)$ and $\left(2\right)$ we have ,
$-(200-h)+50t=h$
or $50t=h+(200-h)$
or $50t=h+200-h$
or $50t=200$
$\therefore t=\frac{200}{50}$
or $t=4s$
Now using equations (1) we get
$h=\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times 4\times 4$
or $h=78.4m$
$\therefore$ The height of meeting point of both the bodies the ground.
$=200-h=200-78.4=121.6m$
Thus both the bodies will meet after $4s$ at the height of $121.6m$ from the ground.