Question

Points $P$, $Q$ and $R$ are in a 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 $PR$?

• A. $\frac{3}{2}$
• B. $\frac{3}{\sqrt{2}+1}$
• C. $\frac{1}{\sqrt{2}-1}$
• D. $\frac{5}{2}$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}-1}$

Let's say $PQ=QR=10meters$

Let's approximate$g=9.8meters/seconds=10meters/seconds$

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

The time to fall distance $PQ$ is:

$t=\sqrt{\left(2h/g\right)}=\sqrt{\frac{2\ast 10}{10}}=\sqrt{\left(2\right)}$

The time to fall distance PR is:

$t=\sqrt{\left(4h/g\right)}=\sqrt{\frac{4\ast 10}{10}}=\sqrt{\left(4\right)}$

The time to fall distance QR is:

$t=\sqrt{\left(4\right)}-\sqrt{\left(2\right)}$

The ratio of the times is$\frac{PQ}{PR}=\frac{\sqrt{2}}{\sqrt{4}-\sqrt{2}}=\frac{1}{\sqrt{2}-1}$