Who can solve this?
I think its not a difficult one, only concept based, but I get stuck with it!

11002

Two inclined planes OA and OB having inclination 30degrees and 60degrees with the horizontal respectively intersect each other at O, as shown in the figure.
A particle is projected from point p with velocity u=10 under root 3 m/s along a direction perpendicular to plane OA. If the particle strikes plane OB perpendicular at Q (take g=10 m/s^2) then,
a) find time of flight from P to Q.(Ans:2sec)
b) The velocity with which the particle strikes the plane OB.(Ans:10m/s)
c) The height h of point P from the ground.(Ans:5m)
d) the distance PQ.(Ans:20m)

actually what I did for the first part is:

I find the time when the initial velocity component becomes perpendicular to final velocity component which comes out to be u/gsin60=2 sec.
but remaining parts..? I have no clue..!

I just want to know which thing can I use to solve the remaining parts. I want to try myself!

Hey administrator! Help me! No one has solved it yet!

Wouldn't part b just be \frac{10}{\sqrt{3}}tan(60)

Actually galactus! I don't get your hint!

I got a hint which is;
just rotate the axes along AO and OB means now AO is y-axis and OB is x-axis, but I am still not able to solve it!

Hi guys! I solved part a as follows;



11158
11159
11160

And part c as follows;

11161

Part b

As the hint says, rotate so that now AO becomes a tall building, and OB is the ground.

Now, you've got a "if I kick a ball horizontally off a building" problem.

You found out that AO is 10m

And the entire thing took 2 seconds.

... which, by my math, should hit at 20 m/s...

... am I missing something here? I haven't checked your math for prev parts... so I'm not entirely sure, but if you're getting the right answers...

Part (b) is solved!

11164

And part (d) is also solved!

11165
11166