Kinematics 4 Question 5

5. Two stones are thrown up simultaneously from the edge of a cliff 240m high with initial speed of 10m/s and 40m/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=10m/s2 )

(2015 Main) (a)

(c)

(b)

(d)

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Solution:

Let us first find, time of collision of two particles with ground in putting proper values in the equation

S=ut+12at2240=10t112×10×t12

Solving, we get the position value of t1=8sec

Therefore, the first particle will strike the ground at 8sec.

Similarly,

240=40t212×10×t22

Solving this equation, we get positive value of t2=12sec Therefore, second particle strikes the ground at 12sec.

If y is measured from ground. Then,

from 0 to 8sec

or

y1=240+S1=240+u1t+12a1t2

Similarly, y2=240+40t12×10×t2

t2y1=30t

(y2y1) versus t graph is a straight line passing through origin

At t=8sec,y2y1=240m

From 8sec to 12sec

y1=0y2=240+40t12×10×t2=240+40t5t(y2y1)=240+40t5t2

Therefore, (y2y1) versus t graph is parabolic, substituting the values we can check that at t=8sec, y2y1 is 240m and at t=12sec,y2y1 is zero.



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