Projectile motion (1 Viewer)

[zeek]

Can someone please post their solution for the projectile motion?! I have no idea how they got L² =2V²t²(1-sin@) -2aVtcos@ + a²
All i got was L² =V²t²cos² -2aVtcos@ + a² after doing Δx=a-Vtcos@ and then squaring both sides.

 

[abcd9146]

ur missing something... i think, well it was kinda the same/different to what i had in the exam, but my memory is all fuzzy @__@

well what i did was use the distance formula and then i eventually got it right... i think pythagoras also worked, but i didn't try that...

you can try it again xD

 

[watkinzez]

Yep, distance formula. Basic expansion and simplification- it's the number of variables that put people off.

 

[nanini]

i had no idea about that question! i still don't see how you do it?!

 

[kakamwamwa]

That was a ridiculous question for question 6!
don't take my word for it but i THINK you had to make the
cos^2x into 1-sin^2x which then became (1+sinx)(1-sinx)
and then somehow the 1+sinx could cancel out...

mmm thats all i can remember. its all a big blur

Does anyone else know the solution?

 

[ianc]

Yeah basically what you had was something like this:

Therefore for the distance, you use pythagoras

L² = (y₂ - y₁)² + (a - x₁)²

where x₁ = vtcos(theta)
y₁ = Vtsin(theta) - (1/2)gt²
y₂ = Vt - (1/2)gt²

if you didn't remember to subtract the y₂, your result will be the same as zeek's. initially i forgot as well...

for ii)

you differentiate L in terms of t, to get a massively ugly thing for dL/dt. You then solve for dL/dt=0 and t will come out to be the thing in the question

Then you substitute t into the formula for L to get that distance (more nasty algebra)

then for iii)

You need to find the time range in which Particle one is ascending

So you differentiate to get y'= Vsin(theta) - gt

Then you find the time when maximum height occurs (ie y'=0)

then you say that this time must be > t=acos(theta)/(2v blah blah)

and you will get V > that thing

 

[3li]

^ exactly wat i did