Asked by Addie
If a projectile has a launching angle of 52 degrees about the horizontal and an initial speed of 18 m/s, what is the highest barrier that the projectile can clear?
Answers
Answered by
Henry
Vo = 18m/s @ 52 deg.
Vo(hor) = 18cos52 = 11.08m/s.
Vo(ver) = 18sin52 = 14.18m/2,up.
Vf^2 = Vo^2 + 2gd = 0 @ max height.
(14.18)^2 + 2(-9.8)d = 0,
201.07 -19.6d = 0,
d = 10.3m. = max barrier that can be
cleared.
Vo(hor) = 18cos52 = 11.08m/s.
Vo(ver) = 18sin52 = 14.18m/2,up.
Vf^2 = Vo^2 + 2gd = 0 @ max height.
(14.18)^2 + 2(-9.8)d = 0,
201.07 -19.6d = 0,
d = 10.3m. = max barrier that can be
cleared.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.