Question

Two stones are thrown up simultaneously from the edge of a cliff $200m$ high with initial speeds of $15m{s}^{-1}$ and $30m{s}^{-1}$. Taking $g=10m{s}^{-2},$ the graph of relative position of the second stone with respect to the first has been shown. The equation of the curved part is

• A. $100+15t-t^2$
• B. $200+30t-5t^2$
• C. $200-15t+5t^2$
• D. $100-30t+5t^2$

Answer 1

Answer: (B)

For the 2nd stone,
$x_2=-15t+\frac{1}{2}(10)t^2$
$\Rightarrow x_2=-15t+5t^2$
Stone will hit the ground when
$x_2=200$ i.e. $200=-15t+5t^2$
$5t^2-15t-200=0$
or $t^2-3t-40=0\Rightarrow (t-8)(t+5)=0$
Rejecting the negative value of
$t\Rightarrow t=8s$ So, $x_2=-15t+5t^2$ for $t\le 8s$
$x_2=200$ for $t>8s$
For the first stone, $x_1=-30t+5t^2$
For $0\le t\le 8s;x_2-x_1=15t$ (straight part) For $t>8s$
$x_2-x_1=200+30t-5t^2$ (curved part)