Question

Two stones are thrown up simultaneously from the edge of a cliff $240m$ high with initial speed $10m/s$ and $40m/s$ respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?
(Assume stones do not rebound after hitting the ground and neglect air resistance, take $g=10m/s^2$)
(The figures are schematic and not drawn to scale.)

• A.
  
• B.
  
• C.
  
• D.
  

Answer 1

For the first ball,
$-240=10t-\frac{1}{2}g{t}^2$
$\therefore 5t^2-10t-240$
$\therefore t^2-2t-48=0$
$\Rightarrow t=8,-6$
The 1st particle will reach the ground in 8 secs.
Upto 8 secs, the relative velocity between the particles is 30m/sec and the relative acceleration is zero..
For the 2nd particle,
$-240=40t-5t^2$
$\Rightarrow t^2-8t-48=0$
$t=12secs$
The second particle will strike the ground in 12 secs.
Option(A): Linear behaviour after 8 secs, hence can be eliminated.
Option(B) shows increase until 12 secs , hence is not correct. after 8 secs, separation will decrease.
Option(c) is the correct one.