projectile launcher assignment (1 Viewer)

[Helstar87]

Hey guys!
I know projectile motion was right back at the start of the Space unit but I have an assignment due this term to determine what effect the launch angle has on the range of flight. I have the assignment part sorted out but our teacher marks the conclusion and analysis part really hard. Could anyone give me some pointers about what certain things should go in the conclusion and analysis? (I dont need specific things for this assignmnent, just general things, that I can then apply to my assignment for example. absolute error ect.)
Thanx heaps!!!

 

[FinalFantasy]

About ur angle thing. Maybe this could help.
Range of projectile is s=ucos θt. Cos 0=1, .: maximum range is when θ=0.(that's when projectile fired horizontally)
Also bcoz range also depends on time of flight, which depends on vertical component of velocity, the bigger usinθ is, da bigger da range, da greater da value of θ, da greater the time of flight.
sin 90=1, so maximum time of flight is when projectile fired vertically.

More stuff:
Time of flight is time taken to reach maximum height and back to origin. i.e s=0
s=ut+ 1\2 at²
Note: u=usinθ
0=utsinθ+ 1\2 at²
0=usinθ+1\2 at
t=-2usinθ\a
Range:
s=utcosθ=ucosθ*(-2usinθ\a)=-2u²sinθcosθ\a=(-u²\a)sin 2θ
Now projectiles launched at 30 degrees and 60 degrees have da same range.
Proof:
s=(-u²\a)sin2θ,
when θ=30, s=(-u²\a)sin 60
when θ=60, s=(-u²\a)sin 120
sin 60=sin 120

 

[wanton-wonton]

An interesting thing about projectile motion is when you fire something at a certain degree at a certain velocity and it covers a certain distance. Then the same thing fired at the same velocity, at an angle complimentary to the original angle will cover the same distance.

E.g. A cannon fired at 30 degrees above horiztonal at 30 metres/sec will cover X distance. The same distance would be covered, if the cannon was fired at 60 degrees above horizontal at 30 metres/sec. This is of course ignoring air resistance. Poof of this is farily lengthy as it involves simultaneous equations (for me anyway, maybe there's a simpler way) but yeah, consider that for your assignment.

 

[Templar]

Proof of this is farily lengthy as it involves simultaneous equations (for me anyway, maybe there's a simpler way) but yeah, consider that for your assignment.

The equation for range involve sin 2x. If one angle is x and the other 90-x, then sin 2x = sin 180-2x = sin 2(90-x) = sin 2(complimentary angle)

 

[FinalFantasy]

Now projectiles launched at 30 degrees and 60 degrees have da same range.
Proof:
s=(-u²\a)sin2θ,
when θ=30, s=(-u²\a)sin 60
when θ=60, s=(-u²\a)sin 120
sin 60=sin 120

lol......................

 

[wanton-wonton]

lol......................

Oh wait....shit...sorry.

EDIT* I didn't read your post, pardon me.

 

[FinalFantasy]

kekeke it's ok.
i was wondering if my posts were invisible lol