18 s
1.A body 'A' is projected upwards with a velocity of 9.8m/s.The second body 'B' is projected upwards with the same initial velocity but after 4 seconds.After how much time will both the bodies meet after the first body is projected?
On using equations of motion i get s=4.9m and t=1 second.But this is not the answer.So how do i continue?
-
UP 0 DOWN 0 0 5
5 Answers
1Hello nihal.
1.How did you solve this problem? I would be glad if you could show me the steps here.
2.Are you sure it is '8 seconds'? Cuz the answer given is 12 seconds.
1is the velocity given right?? i mean that the first body reaches ground after just 2sec. Then how could it meet another body thrown after 4sec??
or that they meet at ground?? if thats the case then the time elapsed will be 6sec (4sec initially and 2sec for the second ball to come back)
infact i am confused with the question... may be something is wrong or i am missing out something!!!
1Folks,
As correctly pointed out by Euclid,the velocity given is incorrect.
Now,I change the question a bit.The particle is projected upwards with a velocity of 98m/s and not 9.8m/s and the rest of the question follows as given above.Now,do we get t=12 seconds?If so,how?
1yes now its done...
s1=s2
98t - 1/2gt2 = 98(t-4) - 1/2g(t-4)2
solving for t, we get t=12s