Question

Two stones are thrown up simultaneously with initial speeds $u_1$ and $u_2(u_2>u_1)$. They hit the ground after $6\text{s}$ and $10\text{s}$ respectively. Which graph correctly represents the time variation of, $\Delta x=(x_2-x_1)$, the relative position of the second stone with respect to the first upto $t=10\text{s}$?
Assume that the stones do not rebound after hitting the ground.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Upto $6\text{s}$
$u_{rel}=u_2-u_1$
$a_{rel}=g-g=0$
Using second equation of motion, we get
$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$
$\Delta x=(u_2-u_1)\times t$
Initially relative acceleration between them is zero, so distance between them will increase linearly.
Hence, $\Delta x$ increases linearly with $t$ upto $6\text{s}$.
From $t=6\text{s}$ to $t=10\text{s}$, first stone is at rest
$a_{rel}=g$
$\Delta x=(u_2-u_1)t-\frac{1}{2}g{t}^2$ (in the downward journey Δx decreases parabolically)
Hence, the correct answer is option (a)