IIT JEE , AIEEE Forum - Physics , General

vikhi72

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)1/2gt^2

d)ut-1/2gt^2

vikhi72

rates will be awarded surely

savvej

i feel that the options are incorrect.i m getting ans as u-g/2(2t-1)

savvej

if the qstn is just t secs then the ans will be ut-1/2gt²

sankydreams

last t secs of its ascent means that last t secs of its going upward
so final velo = 0
accn = -g
time = t
initial velo = v -(- gt) = gt
s = (gt)t - 1/2gt^2
=1/2g^2
hence ans is (c)
rate if this helps u...

elessar_iitkgp

Sankydreams is correct
The time of ascent is u/g
Let the velocity at time u/g - t be v
0 = v - gt
v = gt
Distance travelled in the last t seconds,
s = vt - (1/2)gt² = gt² - (1/2)gt² =(1/2)gt²

netkid07

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