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 Jan, 2011, 10:48: PM

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  | 17th Jan, 2011, 10:11: PM

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