Mav 06 prac exam 2 solutions

[eilatan]

Ok so I'm working through the long answer questions of the MAV 2006 practice exam 2 and have come across a problem in understanding how they have gotten a solution to question 2c)ii (for those playing at home).

I get how they've made the j component of r = 0 to find when the stone hits the ground but to do that they suggest graphing the j component into the calculator. I graphed it as a funtion then as a parametric but I don't get the same shape/answer as they've put in their solutions.

Please help! Am I graphing it wrong?

[fredrick]

i dont see why you need to graph it, but maybe your getting the wrong graph because it is in radian mode?
to get the answer i think you would need to sub in the value for t that u obtained from the j=0 part and sub that into the i component

[eilatan]

Well the solutions suggested I graph it so I tried but it didn't work. My calc was in radian mode when I tried to graph it because of the sine. (It is supposed to be in radian mode yes?)

I just tried to solve for t without graphing it and used the quadratic formula and it worked. Thanks anyhow...but time for the next question/problem I have:

Q4a)
Solution's saying that the arc of the bowl can be modelled as parabola and in actual question asks to show that . Solutions then say that it is obvious that b=0 due to the symmetry of the parabola. Can someone please explain to me how that is obvious?...Thanks.

[fredrick]

Nope, it is suppose to be in degree mode.(You usually use radian mode when dealing with trig fuctions but in this case you have to use degree as you are dealing with degrees)

[eilatan]

OH ok. So if the question gives you the angle in degrees you have to graph it in degree mode...right.

Can someone please answer my 2nd question as mentioned above...?