Question:
With what speed a body be thrown upwards so that the distances covered in the 5th second and $6^{th}$ second are equal?}
With what speed a body be thrown upwards so that the distances covered in the 5th second and $6^{th}$ second are equal?}
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If distance in $n^{th}$ sec = distance in $(n+1)^{th}$ sec, then the time to reach peak is $n$ seconds. $u = n \times g$.
Updated On: May 5, 2026
- $75~m/s$
- $49~m/s$
- $\sqrt{98}~m/s$
- $19.8~m/s$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Distances in consecutive seconds are equal only if the body reaches its maximum height at the end of the first of those two seconds.
Step 2: Meaning
This means the velocity becomes zero exactly at $t=5$ seconds.
Step 3: Analysis
Using $v = u - gt$: $0 = u - (9.8 \times 5)$.
Step 4: Conclusion
$u = 49~m/s$.
Final Answer: (B)
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