Question

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height \(\frac h3\) while going up and coming down respectively.

• A. \(\frac {\sqrt 2-1}{\sqrt 2+1}\)
• B. \(\frac {\sqrt 3-\sqrt 2}{\sqrt 3+\sqrt 2}\)
• C. \(\frac{\sqrt 3-1}{\sqrt3+1}\)
• D. \(\frac 13\)

Answer 1

The Correct Option is B

To solve this problem, we need to analyze the motion of the ball in terms of its kinematics. We are given that the ball is thrown up vertically and reaches a maximum height \( h \). Our objective is to find the ratio of the times when it is at a height of \(\frac{h}{3}\) during its upward and downward journeys.

Step 1: Determine the time taken to reach height \(\frac{h}{3}\) while going up.

When the ball is thrown upwards, it follows the equation of motion:

\(v^2 = u^2 - 2gh\)

At the maximum height \(h\), the final velocity \(v = 0\). Therefore, the initial velocity \(u\) can be expressed as:

\(u = \sqrt{2gh}\)

At a height \(\frac{h}{3}\), the ball's velocity \(v'\) is given by:

\(v'^2 = u^2 - 2g \cdot \frac{h}{3}\)

Substituting the value of \(u\), we get:

\(v'^2 = 2gh - \frac{2gh}{3} = \frac{4gh}{3}\)

\(v' = \sqrt{\frac{4gh}{3}}\)

Now, using the relation \(v' = u - gt_{up}\), where \(t_{up}\) is the time taken to reach \(\frac{h}{3}\) while going up:

\(\sqrt{\frac{4gh}{3}} = \sqrt{2gh} - gt_{up}\)

Upon simplifying, we find:

\(gt_{up} = \sqrt{2gh} - \sqrt{\frac{4gh}{3}}\)

Taking \(g\) common and solving for \(t_{up}\):

\(t_{up} = \frac{\sqrt{2gh} - \sqrt{\frac{4gh}{3}}}{g}\)

After simplifying, we find:

\(t_{up} = \frac{(\sqrt{3} - \sqrt{2})\sqrt{\frac{2h}{3g}}}{2}\)

Step 2: Determine the time taken to reach height \(\frac{h}{3}\) while coming down.

When the ball is descending to height \(\frac{h}{3}\), it comes from the maximum height \(h\) with zero initial velocity, and it accelerates under gravity:

Using the equation:

\(\frac{h}{3} = \frac{1}{2}g(t_{down})^2\)

We can solve for \(t_{down}\):

\(t_{down} = \sqrt{\frac{2\cdot\frac{h}{3}}{g}}\)

After simplifying, we obtain:

\(t_{down} = \sqrt{\frac{2h}{3g}}\)

Step 3: Calculate the ratio:

We now have both times:

• \(t_{up} = \frac{(\sqrt{3} - \sqrt{2})}{2}\sqrt{\frac{2h}{3g}}\)
• \(t_{down} = \sqrt{\frac{2h}{3g}}\)

The ratio of these times is:

\(\frac{t_{up}}{t_{down}} = \frac{\frac{(\sqrt{3} - \sqrt{2})}{2}\sqrt{\frac{2h}{3g}}}{\sqrt{\frac{2h}{3g}}} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\)

Therefore, the correct option is \(\frac {\sqrt 3-\sqrt 2}{\sqrt 3+\sqrt 2}\).