Solved: PHYSICS Question for NEET

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

$\frac{1}{2}gt^2$

$ut - \frac{1}{2}gt^2$

$(u-gt)t$

$ut$

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 .
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).
Calculation: Using the second equation of motion for the downward journey: $h = u't + \frac{1}{2}gt^2$ Substituting $u' = 0$ : $h = 0(t) + \frac{1}{2}gt^2$ $h = \frac{1}{2}gt^2$ .
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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