Two stones are thrown up simultaneously from the edge of a cliff 240   m high with an initial speed of…

Solution

$S_1=10t-\frac{1}{2}g{t}^2$. When $S_1=-240$$\Rightarrow t=8s$. So, at $t=8$ seconds first stone will reach the ground. $S_2=40t-\frac{1}{2}g{t}^2$. Till $t=8\mathrm{ }\mathrm{s}$, $S_2-S_1=30t$. But after $8 \mathrm{s}$ $S_1$ is constant $=-240$. Relative to stone $1$ for ($t_1>\mathrm{8)} \mathrm{s}$ displacements of stone $2$ , $S_2+240$. And at $t=12 \mathrm{s}$ seconds stone will reach the ground. The corresponding graph of the relative position of second stone with respect to first is, 


 

Asked in: JEE Main 2015 (04 Apr)

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