Kinematics: Range doubt
How to go about option (3) & (4)? I have a vague-ish idea it has something to do with radius of curvature and relation from max. height. But how to build on that?
12 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.the frog jumps at what angle? the range depends on the angle right
Yeah, you're right. The radius of curvature is the radius of the circle that passes through the point on the trajectory and is tangent to the trajectory at that point, ie, the path can be locally approximated by the circle. So, the cylinder cross section is that circle for the highest point on the trajectory.
it could've jumped when close to the pipe (high angle), far from the pipe (low angle)
So using that, you can get the horizontal component of velocity
After which it is simple.
Yeah I got:
1. ucos(theta)>sqrt(rg)
2. Time of flight is (a), so usin(theta) can be found from there.
ah, ok i was tripping
But multiplying the time of flight and 1 doesn't give any option
Yeah. The highest point condition must be satisfied
Take FOR of cylinder.
Edit: COM of cylinder
Ignore rotation
Ah.
Oh right that gives (4) ig.
+solved @Opt @Nimboi
Post locked and archived successfully!
Archived by
<@1075951732460376214> (1075951732460376214)
Time
<t:1739161404:R>
Solved by
<@763645886500175892> (763645886500175892), <@717724055217635398> (717724055217635398)