a ball is dropped from a height. if it takes 0.2seconds to cross the last 6m before hittingthe ground . then height from which it was dropped(g=10m/s)
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by newton second eq.
s=ut + .5 gt^2 for last .2 sec u=?,g=10,t=.2,s=6;
then by eq. u=29 m/s
then by third eq.
v^2=u^2+2gs here u=0(initial velocity of ball),v= 29(because it is initial velocity for last .2 sec )
s= hight covered before last .2 sec
s=29*29/20 =84.1 m
total hight 84.1+6=90.1 m
s=ut + .5 gt^2 for last .2 sec u=?,g=10,t=.2,s=6;
then by eq. u=29 m/s
then by third eq.
v^2=u^2+2gs here u=0(initial velocity of ball),v= 29(because it is initial velocity for last .2 sec )
s= hight covered before last .2 sec
s=29*29/20 =84.1 m
total hight 84.1+6=90.1 m
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