question_answer

Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect of the first ? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2) (The figure are schematic and not drawn to scale) [JEE Main Solved Paper-2015 ]

A)                                

B)

C)                                

D)

Correct Answer: A

Solution :

The displacement equation of the first object is\[{{y}_{1}}=10t-\frac{1}{2}g{{t}^{2}}=10t-5{{t}^{2}}\]and the final displacement is \[{{y}_{1}}=-240m\] so, \[-240=10t-5{{t}^{2}}\] \[5{{t}^{2}}-10t-240=0\] \[{{t}^{2}}-2t-48=0\] \[t=\frac{2\pm \sqrt{{{2}^{2}}+4\times 48}}{2}=8\sec \] time taken to reach the ground is 8sec, In the same way second object?s displacement equation is \[{{y}_{2}}=40t-5{{t}^{2}}\]and the taken to reach the ground is \[t=12\]sec. The relative displacement is \[{{y}_{2}}-{{y}_{1}}=30t\] when \[(0<t\le 8\sec )\] \[{{y}_{2}}-{{y}_{1}}=40t-5{{t}^{2}}\] when \[(8\le t\le 12\sec )\] So graph is