First, where did you get this problem? I see why physics students are
confused. How can a soccer ball move NE and W at the same time? Think about
it. I'd prefer to see it as a force problem, with two forces (kicks) acting
on the ball and acting independently, their impulses would cause the
independent velocities. That is, if the westerly force were the only force
then the ball would achieve a final velocity of etc. and if the NE force were
the only force then the ball would achieve a NE velocity of etc.
Having said that, another question. Why are you making it so difficult? Why
not add the vectors?
vector x direction y direction
V1 30.6cos(45) 30.6sin(45)
V2 16.7cos(180) 16.7sin(180)
--------------------------------------------------------
V 4.9 21.6
V is 22.1 m/s at 77.2 degrees.
Bob Carlson
In a message dated 97-10-05 16:07:36 EDT, you write:
<< Greetings everyone. I was wondering if someone could help me out
with a problem that I came across. I was working out a problem and saw
the answer I received did not look right. I tried a few more times and
received the same answer, so I solved it graphically and saw that my
mathematical answer was incorrect. I could not find any mistakes in my
math so I gave the same problem to 3 of our math teachers and they came
up with the same incorrect answer as I did. I was wondering if someone
could help me out.
Here is the problem. A soccer ball is kicked northeast at
30.556m/s at the same time it is kicked westward at 16.667m/s. What is
the resultant velocity and angle of the ball's path?
Here are my steps to solving it. I solved for the first angle
(theta 1) as the angle between the two vectors as 135 degrees. I drew
the parallelogram and solved for the acute angles as 45 degrees (theta
2). I drew the resultant vector and solved its length using the law of
cosine: Resultant = square root (30.556^2 + 16.667^2 -
(2*30.556*16.667*cos 45)) = 22.164m/s
To solve for the angle I chose the angle between the westward
vector and the resultant vector (theta 3). I used the law of sine:
Resultant velocity / sin (theta 2) = northeast velocity / sin (theta 3)
22.164/ sin 45 = 30.556/ sin (theta 3)
theta 3 = 77.12 degrees North of west
If this last answer is true that means that the angle between the
resultant and the northeast resultant is 57.88 degrees. This did not
seem right to me, so I figured that I would use the law of sine to solve
for the angle between the resultant and the northeast vector (theta 4).
Resultant velocity / sin (theta 2) = westward vector / sin (theta 4)
22.164 / sin 45 = 16.667 / sin (theta 4)
theta 4 = 32.12 degrees west of northeast
This last answer is nowhere near the answer I received from the
previous answer. And again, if I take 135 - 32.12 = 102.87, which is
very close to the answer I get when I solved it graphically.
Can anyone see a mistake I made when I got the 77.12 degrees? Is
there a limitation to the law of sines? Yes, I checked to make sure that
my calculators were in the right mode, plus I solved it by hand and kept
getting the same wrong answer (along with the math teachers). Any help
would be most greatly appreciated!