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its indeed 20.32......how did u get it?plz post it.....plz...plz...
gosh dats cool i got it right yeah.........
I would surely tell you....
C assume the initial speed to be 'u'
we got a distace of 8 m to travel before the net which is 2.24 m high. since the original height at which the ball was hit was 3 m, so the difference in height is 3-2.24=.76m. Using horizontal projectile formula i.e. range = u(2h/g)1/2
we get 8=u[(2 x .76)/9.8]1/2
: u=8[9.8/(2 x .76)]1/2
: u=20.32m/s.
Dats all.
Hey I am from Shimla atleast got smeone to interact wid who talks about studies. Could u give me ur e-mail ID ill keep sendin you sme questions.
IT would be kind of you to send me your E-mail ID soon enough. Nd where r u from which skul n which class......
I have been longing to interact with smeone on this site thanks for considering my answer. Could you please give me ur email ID
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