projectile motion (1 Viewer)

[c-duced]

hi...can someone please help?!

Problem: A priate ship 560 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea level, fires balls at initial speed v0 = 82 m/sec.
Question: At what angle from the horizontal must a ball be fired to hit the pirate ship?

 

[Ryan_Kill]

go to the cartesian equation, get it to a quadratic in tan theta (assuming you called the angle with the horizontal thaeta), let it equal 0 and solve.

 

[mojako]

to get the cartesian eqn:
find x = ... t (by integration), then make t the subject
then put it into they eqn for y after you've found it (by integration)
and simplify.

 

[Abtari]

28.1 degrees

Hey, isn't there a formula for range of a projectile (R)

R = (v^2 sin 2x) / g

where R is range, v is the initial velocity of projectile, x is the angle made with the horizontal and g is acceleration due to gravity.

We know v = 82 m/s, R = 560 m, 9 = 10 m/s^2, then we can find the angle x.

According to my calculations, the required angle should be 28.1 degrees.

Does anyone know the answer to this question?

 

[mojako]

using formula is not a good idea
and, say if they Q has 4 marks,
you wont get 4 marks using that formula

 

[Abtari]

Ok Ok, but is my method still correct?

 

[Jase]

Yes, you should probably always derive it from the acceleration.

Whoa.. that was a weird question. i think i may have totally stuffed it up.
I got 61*48` or 39*47` ..

 

[mojako]

Ok Ok, but is my method still correct?

yes

but you defnitely are not allowed to quote that formula in extension 1.
it might be okay in ext2 depending on the nature of the Q.

Hmm I wont bother doing that using cartesian now.. (to check Jase's answer)
but to Abtari:
there will be 2 values of theta.
sin2@ (where @ is theta, strange as it is since it should be alpha )
= something
2@ = "sin^-1(something)" or "pi - sin^-1(something)"

since the domain for theta is
0 <= @ <= pi/2
0 <= 2@ <= pi

 

[Abtari]

Thanks

Oh nooooooo!! I forgot that there are two values because of the domain!!!

Damn! I always keep stuffing up like this!!@#*~

 

[Jase]

er, so whats the answer?

 

[mojako]

er, so whats the answer?

Ok, I did it using the cartesian method
using g = 9.8, I get @ = 63 degrees or 27 degrees.

using g = 10, I get @ = 62 degrees or 28 degrees

Now use the 9.8 as model. My quadratics is:
0 = -228.53 cot^2@ +560 cot@ - 228.53
When g=10 just replace the 228.53 with 233.19

 

[Jase]

well thats weird.. i get one angle right and one angle wrong.
argh im going to do so badly at 3u!

 

[mojako]

hmm..
does this explanation sounds sensible?
you get 39*47` ...
my corresponding angle is 28 degrees
so maybe you wrote 2 as 3
2=3, by inspection