ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Respect on April 06, 2011, 09:23:50 pm
-
hey im confused with this question, because according to my teacher, when we have a "special case projectile motion" case, we cant use the Rmax formula which is Rmax = u^2 sin(2θ) / g .. for such a case..ll its only for symetrical flights.. in that question then? .. how the hell am i suppose to solve this??
-
How do you know that the path is not symmetric? I thought most questions in physics ignores air resistance unless stated in the question otherwise.
-
How do you know that the path is not symmetric? I thought most questions in physics ignores air resistance unless stated in the question otherwise.
clearly their is a ramp ... and its not symetrical. it will go up up and then go down a larger distance . and whjat are u talking about air resistance.. air resistance has nothing to do with this.
-
It's symmetrical, but that equation will find the range between 2 points that are at the same height. If you apply that formula, it will tell you the horizontal distance between the tip of the ramp, and the point where a straight horizontal line frop the tip of the ramp cuts through the drawn trajectory.
You should know how to do all proj. motion questions without using that formula...
Actually, it says to ignore h... so you should be able to apply that equation. Wierd.
-
How do you know that the path is not symmetric? I thought most questions in physics ignores air resistance unless stated in the question otherwise.
clearly their is a ramp ... and its not symetrical. it will go up up and then go down a larger distance . and whjat are u talking about air resistance.. air resistance has nothing to do with this.
umm? the ramp creates the parabolic path and its symmetrical unless there's air resistance. Use the constant accel. equations and you will get the right answer.
P.S. you need to review your proj. motion
-
Ive done it:
You are suppose to use Rmax = u^2 sin(2θ) / g
Rmax is known, 60m.
60= (u^2 sin 60)/10
600=u^2 sin60
600/sin60 = u^2
u=26.3ms-1
-
Ive done it:
You are suppose to use Rmax = u^2 sin(2θ) / g
Rmax is known, 60m.
60= (u^2 sin 60)/10
600=u^2 sin60
600/sin60 = u^2
u=26.3ms-1
wait.. its symetrical if we count the path from the ground, up the ramp.. and down onto another imamginery ramp at the same angle (is that this case) .. or is this a question like..
-
Just read the question properly,(h) is not significant enough to be ignored, thus makes it symmetric.
-
the question stated that the height h can be ignored so yes, it went up a magical ramp at 30 degrees and went down on another ramp at 30 degrees.
-
the question stated that the height h can be ignored so yes, it went up a magical ramp at 30 degrees and went down on another ramp at 30 degrees.
magical ramp.. lol yes i see the stupidity of this question .. they draw a ramp, using ink, which can be seen by ur eye, but write that it doesnt exsist.. extremly confusing and stupid. Such a Neap question.
-
why didnt they just make it a dashed line.. so people who havent dont 1/2 physics would be like oHHH.. yeah..
-
You should know how to do all proj. motion questions without using that formula...
Can someone explain how to do the question without the formula?
-
With projectile motion, the vertical and horizontal speeds of your object are completely independent, atleast in year 12 physics. Also, the horizontal motion is constant. The vertical is only influenced by gravity.
So because it's simplest, we'll look at the horizontal first. Decompose the 30deg vector into its vert and hori components (You can use the notation x and y, or i and j if you wanted) in terms of the velocity which we want to find. It's just a right angled triangle problem.
v(horizontal)=v*cos(30)
Where v is the launch velocity, and v(horizontal) is how fast you are going sideways (to the right - but ignoring the fact that you're also going up)
The horizontal distance we are going is 60m, at a speed of v*cos(30), but how long do we have to make the 60m? That depends on how long it takes to get back to ground level from going up in the air, which depends on the vertical motion. Using vel=dist/time;
v*cos(30)=60/t
Now the vertical is where it gets a tad tricky. First we find the initial vertical velocity,
v(vertical)=v*sin(30)
Since we're going to be using the kinematics equations, maybe we should use 'u' to denote that it is the initial velocity.
u(vertical)=v*sin(30)
So we wanted to find the time... x=ut+0.5att
x=0 (Since we are going up, then returning. The displacement is 0 - remember the kinematics equations work with vectorey things)
a=-10 (The minus is important. The acceleration is due to gravity, DOWN)
u=v*sin(30)
t=? (Lets find it)
And so we substitute:
0=v*sin(30)*t -5tt
5tt=v*sin(30)*t
5t=v*sin(30)
t=[v*sin(30)]/5
Interesting.
Lets go back to my earlier equation, v*cos(30)=60/t. We now know what t is, in terms of v. Lets just substitute and see what happens.
v*cos(30)=60/t, t=[v*sin(30)]/5
v*cos(30)=60/[v*sin(30)]/5
After some maths this becomes;
v=sqrt(300/(sin(30)*cos(30))
v=26.3
(Someone who knows how to do latex might want to fix this) (You have no idea how glad I am that was right.)
In fact having a look at this working, if I knew about trig formulae I think the range equation comes out of that somewhere.
Edit: I checked on wikipedia, turns out that 2cos(theta)sin(theta)=sin(2theta). If you do all the above but all in terms of v, theta and g, you end up with the range equation, x= vvsin(2theta)/g.
So clearly, you should really be using the range equation to do this. But for a lot of problems if you rely on the range equation too much, you could get yourself into trouble.
-
woah, so complicated. But i think i get it. Thanks for the explanation.