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 graphs 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)(The figures are schematic and not drawn to scale.)
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
b
c
d
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
For the second stone time required to reach the ground is given by y = ut -12gt2-240 = 40t-12gt2∴ 5t2-40t-240 = 0(t-12)(t+8) = 0∴ t = 12 sFor the first stone: -240 = 10t-12×10×t2∴ -240 = 10t-5t25t2-10t-240 = 0(t-8)(t+6) = 0T = 8 sDuring first 8 seconds both the stones are in air:∴ y2-y1 = (u2-u1)t = 30tSo, the graph of (y2-y1) against t is a straight line.After 8 seconds,y2 = u2t - 12gt2-240Stone two has acceleration with respect to stone one. Hence graph 3 is the correct description.