Question

Two stones are thrown up simultaneously from the edge of a cliff $240\text{m}$ high with initial speed $10\text{m/s}$ and $40\text{m/s}$ 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 $g=10{\text{m/s}}^2$)
(The figures are schematic and not drawn to scale.)

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Using equation of motion for the first stone,

$-240=10t-\frac{1}{2}g{t}^2$

$\Rightarrow 5t^2-10t-240$

$\Rightarrow t^2-2t-48=0$

$\Rightarrow t=8,-6$

$\therefore$ The 1st stone will reach the ground in $8\text{secs}$.

Upto $8\text{secs}$, the relative velocity between the particles is $30\text{m/sec}$ and the relative acceleration is zero.

For the $\text{2nd}$ stone,

$-240=40t-5t^2$

$\Rightarrow t^2-8t-48=0$

$t=12\text{secs}$

The second stone will strike the ground in $12\text{secs}$.

Now, relative position

$\Delta y=y_1-y_2=40t-\frac{1}{2}g{t}^2-10t+\frac{1}{2}g{t}^2=30t$

$\Delta y=30t$

After $8\text{second}$ stone $1$ reaches ground
$y_1=-240\text{m}$

$\therefore \Delta y=y_2-y_1$

$=40t-\frac{1}{2}g{t}^2+240$

$\therefore$ It will be a parabolic curve after 8 secs.

So, the correct option is $\left(C\right)$.