If a ball were to fall off a 180cm table,how fast would it be going halfway to the ground?
ΔKE = - ΔPE
mv^2/2 - 0=-(mgh-mgH)
mv^2/2=mg(H-h)=mg(H-H/2)=mgH/2
v=sqr(gH)=sqr(9.8•1.8) =4.2 m/s
mv^2/2 - 0=-(mgh-mgH)
mv^2/2=mg(H-h)=mg(H-H/2)=mgH/2
v=sqr(gH)=sqr(9.8•1.8) =4.2 m/s