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Hii sir I have question,a ball is thrown vertically upwards from a bridge with initial velocity of 4.9 m/s .It strikes water after 2 s lf acceleration due to gravity is 9.8 m/s*2 with which velocity does ball strike water
u is given = 4.9 , total time of flight = 2sec applying , S = ut + at2/2 , S = H = ? , a=-g , u=+4.9m/s (upward) putting these in above eq H = 4.9*2 - 9.8(4)/2 = 9.8mV= u + a*tV= 0 + 9.8 *1.5V= 14.7 m/sec
u is given = 4.9 , total time of flight = 2sec
applying , S = ut + at2/2 ,
S = H = ? , a=-g , u=+4.9m/s (upward)
putting these in above eq
H = 4.9*2 - 9.8(4)/2 = 9.8m
V= u + a*t
V= 0 + 9.8 *1.5
V= 14.7 m/sec
14.7m/sSolutionLet h be the height of bridge Above water surface.Let t`,be the time taken to reach highest pointU=4.9m/s. a=-9.8m/s^2•V=u+at0=4.9+(-9.8)t`t`=0.5m/S•v^2=u^2+2as0=4.9^2+2(-9.8)hh=1.225msince total time=2sDownward time =1.5sV=u+atV=0+9.8*1.5v=14.7m/s
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