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

Question

This question for is from topic motion under straight line (free fall). Kindly answer the question simply.

Answer 1

There are 2 different methods to do this problem:

1. theoretical way
2. use sample numbers

Method 1. Theoretical way:

If the intial velocity is 0, the time for a ball to fall a distance h is given by:

t = SQRT(2h/g)

where SQRT is the square root function.

 

I assume that P is at the top of the vertical line and R is at the bottom.

Let's say PQ is h, then the time to fall this distance is:

t = SQRT(2h/g)

Now, PR is twice distance PQ because PQ = QR and:

PR = PQ + QR = h + h = 2h

Then the time to fall 2h is:

t = SQRT(4h/g)

The time to fall QR is the difference of these times:

t_QR = t_PR - t_PQ

 

The subscripts indicate the time to travel the distance.

t_QR = SQRT(4h/g) - SQRT(2h/g)

The ratio of the times is:

t_PR / t_QR = SQRT(2h/g) / [SQRT(4h/g) - SQRT(2h/g)]  

Factor out SQRT(2h/g) in the denomerator:

t_PR / t_QR = SQRT(2h/g) / [SQRT(2h/g)(SQRT(2) - 1)]

Cancel out SQRT(2h/g) in the denominator and numerator:

t_PR / t_QR = 1/ [SQRT(2) - 1] = 1/ (1.414 - 1) = 2.415

Method 2. Use numbers
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Let's say PQ = QR = 10 meters
Let's approximate g = 9.8 meters/seconds = 10 meters/seconds (approximately).

As we saw above, g cancelled out so we can approximate it at no loss in accuracy.

The time to fall distance PQ is:

t = SQRT(2h/g) = SQRT(2*10/10) = SQRT(2)

The time to fall distance PR is:

t = SQRT(4h/g) = SQRT(4*10/10) = SQRT(4)

The time to fall distance QR is:

t = SQRT(4) - SQRT(2)

The ratio of the times is:

SQRT(2) / [SQRT(4) - SQRT(2)] = 1.414 / (2 - 1.414) = 1.414/.586 = 2.415 

This is the same answer as Method 1.