#Projectile motion on an incline
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When the body collides with the plane, it will only have the vertical component of velocity.
As the collision is elastic, this vertical velocity just flips it direction
That means, the object shouldn't reach its original point of projection?
This is what I mean.
Do, we have to consider the fact that the object will roll down the incline to original point (and add this time)
T1= time to go up the incline
T2= for vertically up and down motion
T3= Rolling down
T=T1+T2+T3
why not
Yes, it will roll down?
shouldn’t it bounce back to like exactly where it started or something
How? It landed perpendiculary to the surface of incline
So, at collision, you only have Vy
yeah so it should go back up
is y the direction perpendicular to the incline or the vertical direction
Perpendicular to incline
Like here
yeah ok
the body should literally just retrace its path back to the start though
like the velocity gets negated
why not 
The fact that Vy reverses direction at contact but Vx=0 so net velocity is in Y direction
so the direction is completely flipped right
Only in Y-axis!
As Vx=0 so there is nothing to flip!
it’s 0 in the X-axis
yeah but 0 flipped is 0
Yeah, so vertically up
so it’s completely flipped
In Y axis?
in both axes?????
ok wait can I tell you how I’m imagining the problem
It had no velocity in X-axis, so what is gonna flip
how are you missing my point so hard
if I flip 0 then it is still 0
Can you show a diagram?
so if the velocity was (0, v) before then it’s (0, -v) now
which is the negative of what it was before in both axes
Yes and the direction ofnet velocity is in +Y axis only
I see
0i-vj flips to 0i+vj
purple line is the velocity before collision, blue is after
they are parallel have the same magnitude and are in opposite directions
ok suppose you imagine there’s a video of the ball’s path from where it starts to where it hits the incline
and then as soon as it hits the incline, you play the video backwards
what does that do? it reverses the velocity and doesn’t change the acceleration
Alright, I can imagine that.
Okay, so it retraces its path
so total time=2T where T is the time to go up
should remain to find that ig
your original q was kinda vague there’s like a gazillion variables
not sure if it’s just me who’s reading it wrong
Yeah lmao
$T=\frac{2v}{g \sqrt{1+3sin^2A}}$
I found out T
Agent Krasnov
So $2T=\frac{4v}{g \sqrt{1+3sin^2A}}$
Agent Krasnov
Is the final answer
np :D
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