Exam revision Hard questions

[ghadz7]

Horizontal range = v(squared)sin(2theta)/g

8 = v(squared)sin(30)/10

v = 12.6 m/s 

[shokstar]

Care to explain?

[ghadz7]

When he lands at the same height this formula seems to work...

He's going to have a vertical displacement of zero\
x = ut + 0.5at(squared)

u = vsin(theta)
x = 0
a= -10

vsin(15)t -5t(squared) = 0
t(vsin(15) - 5t) = 0
t = 0 (that's where he started from)
t = vsin(15)/ 5  (the one we are after)

Horiztonal:
u = vcos(15)
x = ut
x= vcos(15)vsin(15)/5
x = 8
8= v(squared) cos(15)sin(15) / 5
v (squared)= 8 * 5 / (cos(15)sin(15))
v = 12.6 m/s

[qshyrn]

Horizontal range = v(squared)sin(2theta)/g

8 = v(squared)sin(30)/10

v = 12.6 m/s  

using the range formula is cheating
horizontal component:
8= vcos15 t   t=8/(vcos15)

Vertical comp:
x=0, a=-10     t=sub in from above= 8/(Vcos15)       u=Vsin15

use x=ut+.5at^2
0=Vsin15*8/(Vcos15)- 5(8/(Vcos15)^2
simplify a little (cross out V's in first bit and turn in into tan 15) then you solve for V

[ghadz7]

It works when the final displacement is 0!

[kenhung123]

can someone show me an example of a banked track question involving friction??

It was mentnioed in another thread.

I've only ever seen friction involved with incline planes, towing, etc. but ive never seen it involved with circular motion.

I hope we don't get one with friction...those are hard..

[kenhung123]

This has friction up the plane..

[Greggler]

is the answer around 37m/s?

[superflya]

yea i got 37. just use

[TyErd]

yup i got 37 as well.

[kenhung123]

its 30.7m/s

[TyErd]

what, no cant be, does it show how they got the answer?

[kenhung123]

..

[m@tty]

Umm, I think they got their formula wrong...

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Unless they have seen something I haven't...

[kenhung123]

Well they say there is frinction up the plane, would that be the reason?