Asked by Kareem
A ball is thrown at 22 m/s at 45 degrees to the horizontal. A 5 foot tall fence is located 100 meters away. Does
the ball make it over the fence?
the ball make it over the fence?
Answers
Answered by
Devron
You can use the range equation to figure this out, first, if the ball even makes it to 100m.
horizontal distance=R=Vi^2*Sin2(Theta)/g
Where
Vi=22m/s
Theta=45
and
g=9.8m/s^2
Solve for R:
R=(22m/s)^2*Sin2(45)/(9.8m/s^2)
R=(22m/s)^2*Sin2(45)/(9.8m/s^2)
R=49.39m
The ball doesn't make it that far.
horizontal distance=R=Vi^2*Sin2(Theta)/g
Where
Vi=22m/s
Theta=45
and
g=9.8m/s^2
Solve for R:
R=(22m/s)^2*Sin2(45)/(9.8m/s^2)
R=(22m/s)^2*Sin2(45)/(9.8m/s^2)
R=49.39m
The ball doesn't make it that far.
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