Asked by Anthony
A football is kicked 66.6 meters. If it is in the air for 5.39 seconds, with what initial velocity was it kicked?
magnitude and direction above horizontal
magnitude and direction above horizontal
Answers
Answered by
Elena
L=vₒ²•sin2α/g = 66.6
t= 2vₒ•sinα/g = 5.39
vₒ = t•g/2•sinα
L=vₒ²•sin2α/g= t²•g²•2•sinα•cosα/g•4•sin²α= t²•g/2•tanα
tanα = t²•g/2•L
α= arctan t²•g/2•L =
=arctan (5.39²•9.8/2•66.6)=64.9°.
vₒ = t•g/2•sinα=5.39•9.8/2•sin 64.9°=...
t= 2vₒ•sinα/g = 5.39
vₒ = t•g/2•sinα
L=vₒ²•sin2α/g= t²•g²•2•sinα•cosα/g•4•sin²α= t²•g/2•tanα
tanα = t²•g/2•L
α= arctan t²•g/2•L =
=arctan (5.39²•9.8/2•66.6)=64.9°.
vₒ = t•g/2•sinα=5.39•9.8/2•sin 64.9°=...
Answered by
Anthony
NEED HELP BAD
Answered by
Anthony
IT IS NOT COMING OUT RIGHT
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