Asked by Anonymous
A projectile is fired at 65.0° above the horizontal. Its initial speed is equal to 37.5 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored.At what time after being fired does the projectile reach this maximum height?
Answers
Answered by
Henry
Vo = 37.5m/s @ 65o.
Yo = 37.5*sin65 = 34 m/s.
Y = Yo + gt.
t = (Y-Yo)/g = (0-34)/-9.8 = 3.47 s.
Yo = 37.5*sin65 = 34 m/s.
Y = Yo + gt.
t = (Y-Yo)/g = (0-34)/-9.8 = 3.47 s.
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.