A ball is thrown up with a velocity of 0.5 m/s .how high will it go before it begins to fall ? how time will it take to reach the height?
Answers
Answered by
2
The peek height it reaches is 0.0125 m
And the time it would take would be 0.05s
Answered by
2
u=0.5 a=-10 v=0
v=u+at
0= 0.5-10*t
t=0.5/10=0.05 second
v²= u²+2as
0 = 0.25+2*-10*s
s= 0.25/20=0.0125m
v=u+at
0= 0.5-10*t
t=0.5/10=0.05 second
v²= u²+2as
0 = 0.25+2*-10*s
s= 0.25/20=0.0125m
Similar questions