Asked by Joseph
A projectile of mass 0.413 kg is shot from a cannon, at height 6.8 m, with an initial velocity vi having a horizontal component of 7.5m/s.
The projectile rises to a maximum height of ∆y above the end of the cannon’s barrel and strikes the ground a horizontal distance ∆x past the end of the cannon’s barrel.
Determine the maximum height the projectile achieves after leaving the end of the cannon’s barrel.
Answer in units of m.
The projectile rises to a maximum height of ∆y above the end of the cannon’s barrel and strikes the ground a horizontal distance ∆x past the end of the cannon’s barrel.
Determine the maximum height the projectile achieves after leaving the end of the cannon’s barrel.
Answer in units of m.
Answers
Answered by
bobpursley
angle:
cosTheta=7.5/vi
time to fall 6.8m:
timein air=sqrt(6.5/g)
horizonal distance=7.5*timeinair
height:
mg*maxheight=1/2 m (vi*sinTheta)^2
solve for max height
cosTheta=7.5/vi
time to fall 6.8m:
timein air=sqrt(6.5/g)
horizonal distance=7.5*timeinair
height:
mg*maxheight=1/2 m (vi*sinTheta)^2
solve for max height
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.