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POINTS P, Q and R are in vertical line such that PQ = QR. A ball at the top most point 'P' is allowed to fall freely. What is the ratio of the times of descent through PQ and QR?

Asked by Ayush Pateria 16th January 2011, 10:48 PM
Answered by Expert
Answer:
Dear Student,
We'll use s = ut + at2/2.
PQ = gtPQ2/2 ... (1)
For QR, u is not zero but, gtPQ.
QR = gtPQ.tQR + gtQR2/2   ... (2)
Since PQ = QR,
gtPQ2/2 = gtPQ.tQR + gtQR2/2 
tPQ /tQR = 2 + tQR /tPQ
 
Let tPQ /tQR = x
x = 2 + 1/x
x - 1/x = 2
x2 -2x - 1 = 0
x = 1 + √2
tPQ /tQR = 1 + √2 = 2.414.
regards,
Team,
TopperLearning.
Answered by Expert 17th January 2011, 10:11 PM
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