Problem: With what speed should a body be thrown upwards so that the distance traversed in the 5th second and 6th second are equal?

Solution:

To solve this problem, we can use the equations of motion for vertical motion under constant acceleration. The key equation we will use is:

\[s = ut + \frac{1}{2} a t^2\]

Where:
- \(s\) is the distance travelled
- \(u\) is the initial velocity
- \(t\) is the time
- \(a\) is the acceleration

We are given that the distance travelled in the 5th second and 6th second are equal. Let's denote this distance as \(d\) for simplicity.



- We need to find the initial velocity (\(u\)) required to traverse the same distance in the 5th and 6th second.
- The body is thrown upwards, so the acceleration (\(a\)) will be equal to the acceleration due to gravity (\(-9.8 \, \text{m/s}^2\)).



We can now apply the equation of motion to find the distance travelled in the 5th and 6th second.

For the 5th second:
\[s_5 = u \cdot 5 + \frac{1}{2} \cdot (-9.8) \cdot (5)^2\]

For the 6th second:
\[s_6 = u \cdot 6 + \frac{1}{2} \cdot (-9.8) \cdot (6)^2\]



Since the distances are equal, we can set \(s_5\) equal to \(s_6\) and solve for \(u\).

\[u \cdot 5 + \frac{1}{2} \cdot (-9.8) \cdot (5)^2 = u \cdot 6 + \frac{1}{2} \cdot (-9.8) \cdot (6)^2\]

Simplifying the equation, we get:

\[5u - \frac{1}{2} \cdot 9.8 \cdot 25 = 6u - \frac{1}{2} \cdot 9.8 \cdot 36\]

Solving for \(u\), we find:

\[u = 49 \, \text{m/s}\]

Therefore, the body should be thrown upwards with an initial velocity of 49 m/s in order to traverse the same distance in the 5th and 6th second.

Happy wala Birthday Mubarakho