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 ?
[JEE mains 2015]
Answers
OPTION B IS CORRECT.
For the first ball,
−240=10t−1/2gt^2
∴5t^2−10t−240
∴t^2−2t−48=0
⇒t=8,−6
The 1st particle will reach the ground in 8 secs.
Upto 8 secs, the relative velocity between the particles is 30m/sec and the relative acceleration is zero..
For the 2nd particle,
−240=40t−5t^2
⇒t^2−8t−48=0
t=12secs
The second particle will strike the ground in 12 secs.
Option(A): Linear behaviour after 8 secs, hence can be eliminated.
Option(B) shows increase until 12 secs , hence is not correct. after 8 secs, separation will decrease.
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Answer:
option (c) is correct graph which represents the time variation of relative position of the second stone with respect to the first
Explanation:
central idea :- concept of relative motion can be applied to predict the nature of motion of one particle with respect to second particle till the first particle strikes ground at a time given by
- -240 = 10t - 1/2×10×t²
- or, t² - 2t - 48 = 0
- or, t = 8, -6
t = -6 ( not possible)
thus , distance covered by second particle with Respect to first particle in 8s is
S₁₂ = (V₁₂)t = (40-10)(8s )
= 30× 8 = 240m
similarly, time taken by second particle to strike the ground is giveb by
- -240 = 40(t) -1/2× 10×(t²)
- or, -240 = 40t -5t²
- or, 5t² -40t -240= 0
- or, t² -8t -48 = 0
- or, t² -12t +4t -48 = 0
- or , t (t - 12) +4(t -12) = 0
- or , t = 12,
T = -4 (not possible)
Thus, After 8s magnitude of relative velocity will increase upto 12s when second particle strikes the ground .
∴ Option (2nd )option (c) is correct