First slide

Vertical projection from ground

Question

If a ball is thrown vertically upwards with speed u, the distance covered during the last ‘t’ seconds of its ascent is:

Easy

A

$u t$
B

$\frac{1}{2} g t^{2}$
C

$u t - \frac{1}{2} g t^{2}$
D

$(u + g t) t$

Solution

Since the motion is symmetric upwards and downwards, the distance covered in last t seconds during ascent is equal to distance covered during 1st t seconds during descent.
Hence, Displacement in last t seconds during ascent= $\frac{1}{2} g t^{2}$

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