If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ seconds of its

Question

If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ seconds of its ascent is

Accepted Answer

1. **Symmetry of Motion:** In the absence of air resistance, time symmetry exists for projectile motion under gravity. The distance covered by an object during the last $t$ seconds of its upward journey is equal to the distance covered by a freely falling object during the first $t$ seconds of its downward journey from the maximum height .
2. **Free Fall Condition:** At the maximum height, the final velocity of the ascent becomes zero, which acts as the initial velocity ($u' = 0$) for the descent (free fall).
3. **Calculation:** Using the second equation of motion for the downward journey:
   $h = u't + \frac{1}{2}gt^2$
   Substituting $u' = 0$:
   $h = 0(t) + \frac{1}{2}gt^2$
   $h = \frac{1}{2}gt^2$.
4. **Alternative Method:** At time $(T-t)$, where $T = u/g$ is the total time of ascent, the velocity is $v = u - g(T-t) = u - g(u/g - t) = gt$. The distance covered in the remaining time $t$ with initial velocity $gt$ and retardation $-g$ is $s = (gt)t - \frac{1}{2}gt^2 = \frac{1}{2}gt^2$.

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