Question

With what speed a body be thrown upwards so that the distances covered in the 5th second and $6^{th}$ second are equal?}

• A. $75~m/s$
• B. $49~m/s$
• C. $\sqrt{98}~m/s$
• D. $19.8~m/s$

Hint:

If distance in $n^{th}$ sec = distance in $(n+1)^{th}$ sec, then the time to reach peak is $n$ seconds. $u = n \times g$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Distances covered in successive seconds are equal only if the body reaches its highest point exactly at the boundary of those two seconds. In this case, the body must reach the peak at $t = 5$ seconds so that it ascends in the 5th second and descends in the 6th second.
Step 2: Key Formula or Approach:
At the highest point, velocity $v = 0$. Using $v = u - gt$.
Step 3: Detailed Explanation:
For the distance in the 5th second to equal the distance in the 6th second, the time of ascent must be 5 seconds. \[ 0 = u - g(5) \] \[ u = 5g \] Taking $g = 9.8 \text{ m/s}^2$: \[ u = 5 \times 9.8 = 49 \text{ m/s} \]
Step 4: Final Answer:
The initial speed should be 49 m/s.