Kilbaha Exam 2 08

[nerfsdacier]

Hey everyone,

I'm just a bit confused with a few mcq from this exam: (see attachment)

So with Q10: the answer is E, which i get, but I think the range for D is also incorrect. Isn't the range for D: [a, infinity) ?

Q14: Answer is A which I understand, but if you transpose all the forces, I thought they make a right angled triangle, so I thought D is also correct?

Q18: They say the answer is C, but I don't quite agree with them.
They said that, u = 0, a =9.8, t = 8s, therefore, h = 0.5 x 9.8 x 8 x 8  = 313.6m
But, since the hot air balloon is accelerating upwards, shouldn't the stone have an initial velocity upwards as well?
so i went:
initial velocity = -sqrt(2h) (take down as positive)
h = -sqrt(2h)x 8 + 0.5 x 9.8 x 8 x8
for which i get h = 167m, which is none of the choices.

Thank you for your help

[Conic]

Q14: See attached.

[nerfsdacier]

Thanks for the speedy reply I'm still confused though:

Q10: what if x = 1/(2b).. then wouldnt the denominator = sqrt(3/4)? and so the fraction would be greater than a? I feel like I'm missing something really obvious in this one... I sketched it subbing a =1 =b , and get the range to be [1, infinity) ?
Q14: how do you know the angle between g and T1 = 90degrees? wouldnt it be 46 degrees?

[Conic]

Q18: I'm pretty sure it does not have initial velocity, so you were right:



Q14: The angle between T2 and g is given by 90-46, the angle between T1 and g is given by 180-((90-46)+(90-44)) and the angle between T1 and T2 is given by 90-44. It's one of the basic triangle of forces tricks I learned in class.

Q10: I'm not sure about this question, so I can't be of much help.

[BubbleWrapMan]

Answer to Q10 is D, since as x approaches 1/b, f(x) approaches positive infinity, so its range is incorrect