Asked by nam maurer
A projectile was launched 64° above the horizontal, attaining a height of 10 m. What is the projectile's initial speed?
Answers
Answered by
drwls
(Average vertical velocity) * (time of flight to max height) = Maximum height H
(Vo*sinA/2)*(VosinA/g) = H
Vo^2*sin^2A/(2g) = H
Solve for Vo
(Vo*sinA/2)*(VosinA/g) = H
Vo^2*sin^2A/(2g) = H
Solve for Vo
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.