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 2015]

A)     

B)

C)    

D)

Correct Answer: A

Solution :

[a] 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