CSSA 2004 question 18 (1 Viewer)

[fashionista]

has anyone done this question? im so confused, i was saying that the observer on the platform would see the lightening strike the front of the train first and then the back while the person inside the train would see it strike at the same time.

im wrong arent i?

 

[wind]

Isn't this like a role-reversal of Einsteins train thing?

Instead of the passenger seeing the lightning simultaneously strike the carriage at either end, it's the stationary observer that sees this happening.

The passenger sees the front of the train being struck before the end.

Both the observer and lightning are in the same frame of reference. Or am I wrong? lol

 

[helper]

Yes.
There is a big discussion of this on another thread in space.

 

[mojako]

Yes.
There is a big discussion of this on another thread in space.

LOL
so thats all about CSSA

 

[wind]

I hate the CSSA paper! Stupidest thing ever!

 

[fashionista]

it really is!! i cant do this projectiles question, question 19(b)
PLEASE HELP ME ITS SHITTING ME TO TEARS

 

[wind]

Was that the huge one with the rocket or something?

 

[fashionista]

no no, it just says its a projectile being fired, but its a four marker and im screwed if i know how to do it. it goes, a projectile is fired at an angle of 30 degrees to the horizontal from a tower 20 m high and takes 10 secs to hit the ground. calculate the initial speed of the projectile.

i cant do it

 

[helper]

Vertical
y=u_yt + 0.5 a_yt²

-20=usin30*10+0.5*(-9.8)*10²
-20+490=5u
u=470/5
=94 m/s at angle 30 degree above horizontal.

Method correct. Check maths

 

[mojako]

This is if the projectile is fired 30 degrees above horizontal

find max height... using the syllabus notation and language technique of symbolism,
u[y]=u*sin30=0.5u
we have v[y]^2=u[y]^2+2a[y]delta(y) so putting v[y]=0 and a=-9.8,
the value of delta(y) gives the max height above the tower.

you can then use delta(y)=u[y] t + 1/2 a[y] t^2 to find t using quadratic formula
call this t[1] which is the time it takes to rise up


now consider the motion downwards
max height = 20 + delta(y) from the ground
this will be the height it has to travel down

use delta(y)=u[y] t + 1/2 a[y] t^2 but with delta(y)=the max height above ground, a=+9.8, u[y]=0 to find t.. call this t[2] and this is the tme taken to fall down and hit the ground

t[1]+t[2]=10 and you should be able to work out u now

note: I havent actually done this on paper so hope it works.