a ball is dropped from a hight if it takes 0.2 sec to cross the last 6m before beating the ground. Find the hight from these.Which was dropped? g=10m/s square
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hey dear your answer is here..
t=0.2 sec
u=?
s=6m
g=10m/s2
s=ut + (gt^2)/2
6=u×0.2+(10×(0.2^2))/2
so u = 29m/s
now this u is final velocity when we drop it from height x m. and u=0,v =29m/s
2gh=v^2-u^2
2×10×x=29^2-0
x=42.05
so now total height= 42.05+6= 48.05m
hope it helps you dear
t=0.2 sec
u=?
s=6m
g=10m/s2
s=ut + (gt^2)/2
6=u×0.2+(10×(0.2^2))/2
so u = 29m/s
now this u is final velocity when we drop it from height x m. and u=0,v =29m/s
2gh=v^2-u^2
2×10×x=29^2-0
x=42.05
so now total height= 42.05+6= 48.05m
hope it helps you dear
priyalakra93:
thanks
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