00:01
Okay, so this is the graph i drew for this question.
00:05
So we start rolling from point a and it reached point b, and at point b, you start flying with the speed b here.
00:13
Okay? and when it's flying, it starts with an angle of 35 degrees above the horizontal surface.
00:22
So according to energy conservation law, we know that we'll have 1 .5 -i -omega -square plus 1 -5m -v -a squared plus m -g -h -a is equal to 1 -5 -5 -i -o -o -m -m -m -a.
00:33
B to the power 2 plus 1 1 .5 mvb to the power 2 plus m ghb.
00:37
So 1 1 .5 omega a square here is the rotational energy at point a.
00:46
And 1 half mv a square here is the kinetic energy at point a.
00:50
And mgha is the potential energy at point a.
00:57
And when i have omega b square here is the rotational energy at point b and the 1 half mvb square here.
01:05
Here is the kinetic energy at point b, and mghb is the gravitational potential energy at point b.
01:11
So we know vhb equal 0, which means that omega -a should be equal 0 as well, since omega -a is equal to va over r.
01:17
Since we know that initially it was at rest, so therefore we have mgha is equal to 1 1ā2 i -o -b squared plus 1 -5 mvb squared plus mghb.
01:26
So we know hb should be equal 0 based on the graph here...