If a ball is thrown vertically upwards with speed u , the distance covered during the last t seconds of its acsent is ? (a) ut (b)1/2gt2 (c) ut-1/2gt2 (d) (u+gt) t
Let H be the total height and T be the time of ascent H = uT-1/2gT2 Let the distance covered by ball in (T-t) sec by y. y= u(T-t)- 1/2g(T-t)2 Distance covered by ball in last t sec is h=H-y = ( uT-1/2 gT2)- [ u(T-t)-1/2g(T-t)2] h=uT-1/2gT2-uT+ut+1/2gT2+1/2gt2-gTt h= ut+1/2gt2-gTt put T= u/g in above equation we get h= ut+1/2gt2-gt(u/g) h=1/2gt2
