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NEET PHYSICSMedium

If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first three seconds, the time of the travel is:

A

6 sec

B

5 sec

C

4 sec

D

3 sec

Step-by-Step Solution

  1. Analyze the Motion: The body starts from rest (u=0u=0) and undergoes free fall with acceleration gg. Let the total time of flight be tt seconds. The 'last second' corresponds to the tt-th second.
  2. Distance in First 3 Seconds: Using the equation s=ut+12at2s = ut + \frac{1}{2}at^2: S3=0+12g(3)2=9g2S_3 = 0 + \frac{1}{2}g(3)^2 = \frac{9g}{2} .
  3. Distance in Last (tt-th) Second: The distance covered in the nn-th second is given by Dn=u+a2(2n−1)D_n = u + \frac{a}{2}(2n - 1). Here n=tn=t: Dt=0+g2(2t−1)D_t = 0 + \frac{g}{2}(2t - 1)
  4. Equate and Solve: The problem states Dt=S3D_t = S_3. g2(2t−1)=9g2\frac{g}{2}(2t - 1) = \frac{9g}{2} 2t−1=92t - 1 = 9 2t=10  ⟹  t=5 s2t = 10 \implies t = 5 \text{ s}
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