Step 1: Understanding the Question:
When a body is thrown vertically upwards, it decelerates until it reaches zero velocity at its highest point, then accelerates downwards.
The distances in consecutive seconds are equal only if the highest point is reached exactly at the end of the first of these seconds (symmetric motion around the peak).
In this case, the distance covered during the upward part of the 5th second must equal the distance covered during the downward part of the 6th second.
Step 2: Key Formula or Approach:
For distances in the \(5^{th}\) and \(6^{th}\) seconds to be equal, the time to reach the highest point must be exactly at the midpoint of these intervals, which is at \(t = 5\) s.
Formula for time to reach max height: \(t = u/g\).
Step 3: Detailed Explanation:
Distance in \(n^{th}\) second is \(s_n = u - \frac{g}{2}(2n - 1)\).
For the \(5^{th}\) second (\(n=5\)): \(s_5 = u - \frac{g}{2}(2 \cdot 5 - 1) = u - 4.5g\).
For the \(6^{th}\) second (\(n=6\)): \(s_6 = u - \frac{g}{2}(2 \cdot 6 - 1) = u - 5.5g\).
Note: Distances are scalar quantities. In the \(6^{th}\) second, the body is moving downwards, so the displacement is negative, but we equate magnitudes of distance.
Alternatively, and more intuitively, the body reaches maximum height at \(t = 5\) s. At this point, velocity is zero.
If maximum height is reached at \(t = 5\) s, then:
\[ v = u - gt \]
\[ 0 = u - g(5) \]
\[ u = 5g \]
Taking \(g = 9.8\) m/s\(^2\):
\[ u = 5 \times 9.8 = 49 \text{ m/s} \]
Let's verify: In 5th second it travels from \(t=4\) to \(t=5\) (upwards). In 6th second it travels from \(t=5\) to \(t=6\) (downwards). By symmetry, these distances are equal.
Step 4: Final Answer:
The initial speed required to make distances in the 5th and 6th second equal is 49 m/s.