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

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

(a) (u + gt)t

(b) ut

(c) \(\frac{1}{2}\)\(gt^2\)

(d) ut − \(\frac{1}{2}\)\(gt^2\)

Answer :  (c) \(\frac{1}{2}\)\(gt^2\)

v = u + at

0 = \(u_2\) - gt

\(u_2\) = gt

s = ut + \(\frac{1}{2}\)\(gt^2\)

= \(u_2\)t + \(\frac{g}{2}\)\(t^2\)

= \(\frac{1}{2}\)\(gt^2\)