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Two stones are thrown up simultaneously  from the edge of a cliff 240 m high with initial speed of 10 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 to the first?  (Assume stones do not rebound after hitting  the  ground  and  neglect  air resistance, take g=10 m/s2


  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)


Answers (1)


As we discussed in

2nd equation or Position- time equation -

s= ut +\frac{1}{2}at^{2}

s\rightarrow Displacement

u\rightarrowInitial velocity


t\rightarrow time



 For stone 1 y_{1}= 10t-\frac{1}{2}gt^2


For stone 2     y_{2}= 40t-\frac{1}{2}gt^2


\Delta y=y_1-y_2= 40t-\frac{1}{2}gt^2-10t+\frac{1}{2}gt^2=30t

\Delta y =30t

After  8 second stone 1 reaches ground

y_1 =-240m

\therefore \Delta y= y_2-y_1 

    = 40t-\frac{1}{2}gt^2+240

therefore it will be a parabolic curve.

Correct option is 3.

Option 1)


Option 2)


Option 3)


Option 4)


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