That's the way I was picturing it.
Is it similar to a plane flying in crosswind?
That's the only thing I can think of :X which is in the relative velocity section.
The example/figure/diagrams also look similar.
I'm not too strong in the relative velocity section :(
Can anyone give some more...
I know my angle A is 60 degrees because the camel walked 30 degrees south of west and then turned straight ahead north, thus creating a 60 degree turn.
Correct me I'm wrong please -_-
using other law of cos:
solving for B angle:
252= (5(31)1/2)2 + 302 - 2 (5(31)1/2(30)cosB
625 = 775 + 900...
Alrighty, looks like I had computational error.
0.5 = 3 -(9.8*9/(Vo^2*cos^2(45))/2
11.025/ [Vo^2cos^2(45)]= 2.5
11.025/2.5 = Vo^2 * cos^2(45)
4.41 = [Vo^2]/2
8.82 = Vo^2
2.9698 m/s = Vo
how about that?
Ooo yeah...which is why we use law of cos!
a2=252+302-[(2*25*30)cos60o]
a2=625+900-[(2*25*30)cos60o]
a2=1525 -1500cos(60o) = 775
a=5*(31)1/2
so that's the magnitude of the a vector?
now direction would be 30o north of west with a distance of 5*(31)1/2km?
How would I answer the...
R= [Vo^2 * sin2Ao] / g
was the key to solving this problem then?
R = range of projectile = maximal = 8.
sin(2Ao) = maximal and will never be greater than 1 so it has to be = 1 which is 90 degrees or 2 * 45 degrees ?
Could you explain that a little more?
Do I assume that the 30km walk north completes a 90 degree right triangle?
So logically, the camel could have went straight west and hit the 2nd oasis?
But then wouldn't the hypotenuse = 25km, which is less than the base = 30km?
R= [Vo^2 * sin2Ao] / g
where R = range = x = 8.8m
8 = [(10.2^2)*sin2Ao]/ 9.8m/s
8/9.8 = 10.2^2 *sin2Ao
[8/9.8]/(10.2^2) = sin2Ao
(sin2Ao = 2sinAocosAo ?)
[(8/9.8)/(10.2^2)]/2 = sinAoCosAo ?
Um...I don't know what to do from there...?
or if h was the given maximum height...
Ok, so I've drawn the diagram for this.
It looks like this is solved with trig?
I get a right triangle with the top as 2.6m. and the lower 45 degree angle as 2.1m.
So the height of the triangle would be 0.5m and the base would be 3m.
Then, what does the hypotenuse equal?