Question

Two stones are thrown up simultaneously from the edge of a cliff $200\, m$ high with initial speeds of $15 \,ms ^{-1}$ and $30 \,ms ^{-1}$. Taking $g=10\, ms ^{-2},$ 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. $100+15 t-t^{2}$
• B. $200+30 t-5 t^{2}$
• C. $200-15 t+5 t^{2}$
• D. $100-30 t+5 t^{2}$

Answer 1

Answer: (B)

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