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NEET Physics QUESTION #7301
Question 1

A ball is thrown vertically upward with initial speed $u$. The distance covered during the last $t$ seconds of its ascent is:

  • $ut$
  • $\dfrac{1}{2}gt^2$✔️
  • $ut - \dfrac{1}{2}gt^2$
  • $(u+g)t$
Correct Answer Explanation

By time-reversal symmetry: the last $t$ seconds of ascent is equivalent to free fall from rest for $t$ seconds.

Distance $= \dfrac{1}{2}gt^2$