Question

If a particle is thrown vertically upwards, find ratio of time taken to traverse lower half to upper half of journey to the
highest point.​

Answer 1

Answer:

$(\sqrt{2}-1):1$

Explanation:

In the question,

Let the height to which the particle goes is = h metre.

and,

Velocity of the throw of the particle = u m/s.

So,

Using the Equations of Motion,

$v^{2}-u^{2}=2as\\0^{2}-u^{2}=2(-g)h\\u=\sqrt{2gh}$

So,

At the half of the distance the speed of the particle is given by,

$v^{2}-u^{2}=2as\\v^{2}-2gh=2(-g)(\frac{h}{2})\\v^{2}=2gh-gh=gh\\v=\sqrt{gh}$

Therefore,

Again using the law of motion for the first half,

$v=u+at\\\sqrt{gh}=\sqrt{2gh}-g(t)\\gt=\sqrt{gh}(\sqrt{2}-1)\\t=\sqrt{\frac{h}{g}}(\sqrt{2}-1)$

Also,

For the second half,

$v=u+at\\0=\sqrt{gh}-gT\\T=\sqrt{\frac{h}{g}}$

Therefore, the ratio of the time taken for covering the first half of the distance to the time take in covering the second half of the distance is given by,

$\frac{t}{T}=\frac{\sqrt{\frac{h}{g}}(\sqrt{2}-1)}{\sqrt{\frac{h}{g}}}\\\frac{t}{T}=(\sqrt{2}-1):1$

Therefore, the ratio is,

$(\sqrt{2}-1):1$