Question

when and where the ball will meet the elevator.

Answer 1

Let the ball and the elevator will meet after time $t$ at the level where the ball has an upward displacement $x$.

For ball : $u=18$ $m/s$ $a=-g=-9.8$ $m/s^2$

using $S=ut+\frac{1}{2}a{t}^2$

$\therefore$ $x=18t-\frac{9.8}{2}{t}^2$

$⟹$ $x=18t-4.9t^2$ ..............(1)

For elevator : $(7+x)=vt$

$\therefore$ $7+x=2t$

OR $7+18t-4.9t^2=2t$ [using (1)]

OR $4.9t^2-16t-7=0$

$⟹$ $t=\frac{16\pm \sqrt{{16}^2+4\left(4.9\right)\left(7\right)}}{2\times 4.9}=3.65$ s OR $-0.39$ s (Not possible)

$\therefore$ $x=2\left(3.65\right)-7=0.3$ m

Thus the ball and the elevator will at $12.30$ m level after $3.65$ s