Projectile Motion-help (1 Viewer)

[chocolate_lover]

Ok, for the syllabus dot point: "perform a first-hand investigation, gather information and analyse data to calculate initial and final velocity, maximum height reached, range and time of flight of a projectile for a range of situations by using simulations, data loggers and computer analysis"

The data I've got is:
the initial velocity/"u" = 4.36851 ms^-1
angle = 33
measured range = 174.13cm

So I need to work out:
the maximum height
time of flight, and
calculated range

What I'm stuck on, is which formula to use to work these out.
Thanks

edit: The landing point of the object is level with the point of launch. (Thanks airie)

 

[airie]

At max height, v_y = 0, ie. v_y = u_y + a_yt = 0. And a_y is just gravity and equals -9.8ms^-2, u_y = u sin(theta), so you can work out t. Then sub this value into the formula delta(y) = u_yt + 0.5a_yt² to get delta(y).

I'm assuming the landing point of the object is level with the point of launch, since you didn't say that you started a distance above the ground or anything

So delta(y) = 0 ie. u_yt + 0.5a_yt² = 0.

Since u_y = u sin(theta), you've got u and theta there, and as before a_y = -9.8ms^-2, you can calculate the time of flight, t.

Sub this t into delta(x) = u_xt where u_x = u cos(theta) to get the calculated range.

 

[SoulSearcher]

It really doesn't help me that I only know the extension 1 maths methods of figuring out these sorts of stuff

 

[airie]

You're welcome chocolate_lover

It really doesn't help me that I only know the extension 1 maths methods of figuring out these sorts of stuff

Tell me the ext 1 maths method of doing it I'm yet to do the topic in maths at school

 

[twilight1412]

well in yr 11 you learnt in physics that the
derivative or displacement is velocity
derivative of velocity is acceleration ><
so you go from there

i have also yet to learn both methods ^^
my skools doing motors and generators first for some reason

@_@

 

[alcalder]

Can I just say now, there is no right or wrong way of doing these questions. You acn use the Physics equations OR you can use Ext 1 Maths to do it. I always preferred the Ext 1 maths way because you can derive equations that take into account initial conditions and whether you start on top of a cliff or whatever.

Don't feel like you are "cheating" using the calculus.

 

[airie]

Can I just say now, there is no right or wrong way of doing these questions. You acn use the Physics equations OR you can use Ext 1 Maths to do it.

I know, but I'm just curious about the ext 1 maths way to do it, since I haven't done it at school

I always preferred the Ext 1 maths way because you can derive equations that take into account initial conditions and whether you start on top of a cliff or whatever.

You can still do that - if you're starting off on a cliff x metres high, and you're looking at when the object lands on ground level, just put delta(y) = -x

Don't feel like you are "cheating" using the calculus.

Heh, cheating... I just think that there should always be a method that doesn't involve calculus, somehow XD

 

[chocolate_lover]

Thanks, I think I understand most of it now.
Also, could someone post up the Extension 1 way of working it out?

Btw, i worked out:
the maximum height = 0.85344
time of flight, and = 0.48
calculated range = 1.78(2dp)

If anyone else is bothered to work this out, could u please tell me what u got?

 

[alcalder]

OK to do it the Ext 1 way, first take the initial velocity v at an angle θ to the horizontal and resolve it into vertical (y-direction) and horizontal (x-direction) components.

INTIAL CONDITIONS
at t=0
u_y = vsinθ
u_x = vcosθ

If you start on the horizontal, then at t=0, x=0 AND y=0
If you start on a cliff, say 100m off the ground, at t=0, x=0 and y=100

Up is always + and down is - direction.

DERIVING THE EQUATIONS
Work on the x and y directions on separate sides of the page.

The only force acting on an object in gravity and gravity acts only in the y direction and it acts down. Thus:

a_y= -g (where g=9.8m/s)

Integrate to get
v_y = -gt + C (C a constant)

use intial conditions to find C, when t=0, v_y = vsinθ

THUS
v_y = -gt + vsinθ

Integrate to get
y = -gt²/2 + vtsinθ + C (C another constant)

Use initial conditions to get (assuming here we are on the horizontal), when t=0, y=0

THUS
y = -gt²/2 + vtsinθ

[If we were on the cliff, then
y = -gt²/2 + vtsinθ + 100
Which is why this method is better for these sorts of weirder questions because you then don't need to think too hard about stuff when using the equations.]

NOW, to the x-direction. There is no acceleration in the x-direction. Thus:

a_x = 0

Integrate
v_x = C = vcosθ

Integrate again
x = vtcosθ + C
Use initial conditions, t=0, x=0

THUS
x = vtcosθ

Your equations are then:
y-direction
a_y= -g (where g=9.8m/s)
v_y = -gt + vsinθ
y = -gt²/2 + vtsinθ OR y = -gt²/2 + vtsinθ + 100 (depending on where you start in the vertical direction)

x-direction
a_x = 0
v_x = vcosθ
x = vtcosθ
SO
Max height when v_y = 0
Max distance when y=0

And so on.

I hope this helps.

 

[SoulSearcher]

Always fun to do it that way

 

[airie]

OK to do it the Ext 1 way, first take the initial velocity v at an angle θ to the horizontal and resolve it into vertical (y-direction) and horizontal (x-direction) components.

INTIAL CONDITIONS
at t=0
u_y = vsinθ
u_x = vcosθ

If you start on the horizontal, then at t=0, x=0 AND y=0
If you start on a cliff, say 100m off the ground, at t=0, x=0 and y=100

Up is always + and down is - direction.

DERIVING THE EQUATIONS
Work on the x and y directions on separate sides of the page.

The only force acting on an object in gravity and gravity acts only in the y direction and it acts down. Thus:

a_y= -g (where g=9.8m/s)

Integrate to get
v_y = -gt + C (C a constant)

use intial conditions to find C, when t=0, v_y = vsinθ

THUS
v_y = -gt + vsinθ

Integrate to get
y = -gt²/2 + vtsinθ + C (C another constant)

Use initial conditions to get (assuming here we are on the horizontal), when t=0, y=0

THUS
y = -gt²/2 + vtsinθ

[If we were on the cliff, then
y = -gt²/2 + vtsinθ + 100
Which is why this method is better for these sorts of weirder questions because you then don't need to think too hard about stuff when using the equations.]

NOW, to the x-direction. There is no acceleration in the x-direction. Thus:

a_x = 0

Integrate
v_x = C = vcosθ

Integrate again
x = vtcosθ + C
Use initial conditions, t=0, x=0

THUS
x = vtcosθ

Your equations are then:
y-direction
a_y= -g (where g=9.8m/s)
v_y = -gt + vsinθ
y = -gt²/2 + vtsinθ OR y = -gt²/2 + vtsinθ + 100 (depending on where you start in the vertical direction)

x-direction
a_x = 0
v_x = vcosθ
x = vtcosθ
SO
Max height when v_y = 0
Max distance when y=0

And so on.

I hope this helps.

...Why am I not surprised that the maths ext 1 method would involve all this calculus...

 

[alcalder]

It's not a lot of calculus and the integrals are really really simple ones

But, hey, Ext 1 is all about Calculus, isn't it ? Ah the joys of calculus.:rofl: