Carrotsticks
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Regarding this whole 'MIF hating' agenda, please stop it now. It is getting stupid.
Applications of Calculus to the Physical World (1 Viewer)
- Thread starter FDownes
- Start date
davidgoes4wce
Well-Known Member
Having come across 4 students in the past year doing prelims and HSC. I found 3 of the 4 students using Grove as their school text book.
Not sure what the percentage state wide would be for using MIF but I still reckon it would be high proportion. I had a maths tutor when I was in Year 11 and he had a rule. 'If you study from just 1 textbook and do all the questions your guaranteed around 65%, if you do at least 2 books you can get around 85%, if you do 3 books your around the 90% mark, if you do 3 books+past papers well and thoroughly you can push towards 95% mark'.
I think with maths that is a fair estimate to go by.
Not sure what the percentage state wide would be for using MIF but I still reckon it would be high proportion. I had a maths tutor when I was in Year 11 and he had a rule. 'If you study from just 1 textbook and do all the questions your guaranteed around 65%, if you do at least 2 books you can get around 85%, if you do 3 books your around the 90% mark, if you do 3 books+past papers well and thoroughly you can push towards 95% mark'.
I think with maths that is a fair estimate to go by.
enigma_1
~~~~ Miss Cricket ~~~~
Ok
Lol what. That's an invalid generalisation. It would depend on a whole range of factors such as which questions you do, which textbooks you use, how many useful questions you do etc. Tbh even if you use whatever textbook and still do plenty of past papers, you can do really well. Textbook questions can be quite different to past papers and it's best to expose yourself to exam questions since that's what you'll be tested on in the end.
K enough derailing, I'm out
davidgoes4wce
Well-Known Member
FDownes Q 12. (Grove E6.11 Q13) A gun is aimed at a target on the ground 150m away. If the initial velocity is 125ms-1, find the angles at which the gun could be fired to reach the target (use g = 10ms-2)
Vertical Components
ÿ=-10
ẏ=∫ -10 dt= -10t + C
when t=0, ẏ=125 sin θ
C=125 sin θ
ẏ=-10t + 125 sin θ
y=∫-10t + 125 sin θ dt
y=-5t^2 + 125 t sin θ + C
t=0 , y=0
C=0
y=-5t^2+125 t sin θ
Horizontal Components
ẍ=0
ẋ=∫ 0 dt= 0t +C
when t =0, ẋ=125 cos θ
C=125 cos θ
ẋ=125 cos θ
x=∫ 125 cos θ dt
x=125 t cos θ +C
t=0, x=0
C=0
x=125 t cos θ
We now know our vertical and horizontal motion equations.
We let x=150 and y=0
a) 150=125 t cos θ
b) 0=-5t^2+125 t sin θ
Rearranging a) to get in terms of 't'
t=150/(125 cos θ)
0=-5[(150/125 cos θ)]^2 + 125 [150/(125 cos θ)] sin θ
This simplifies to
0=-5[1.44/(cos θ)^2]+150 tan θ
0=-7.2[1+(tan θ)^2]+150 tan θ
0=-7.2-7.2 (tan θ)^2+150 tan θ
Rearranging,
7.2 (tan θ)^2-150 tan θ+7.2=0
Using the Quadratics roots formula
tan θ=[-(-150) ± sqrt[22500-4(7.2)(7.2)]]/ 2 (7.2)
tan θ= (150 ±149.3072)/14.4
θ=87.25 degrees or 2.75 degrees
OR
θ=87°15' or 2°45' (converted into degrees and minutes as required in the solutions)
Vertical Components
ÿ=-10
ẏ=∫ -10 dt= -10t + C
when t=0, ẏ=125 sin θ
C=125 sin θ
ẏ=-10t + 125 sin θ
y=∫-10t + 125 sin θ dt
y=-5t^2 + 125 t sin θ + C
t=0 , y=0
C=0
y=-5t^2+125 t sin θ
Horizontal Components
ẍ=0
ẋ=∫ 0 dt= 0t +C
when t =0, ẋ=125 cos θ
C=125 cos θ
ẋ=125 cos θ
x=∫ 125 cos θ dt
x=125 t cos θ +C
t=0, x=0
C=0
x=125 t cos θ
We now know our vertical and horizontal motion equations.
We let x=150 and y=0
a) 150=125 t cos θ
b) 0=-5t^2+125 t sin θ
Rearranging a) to get in terms of 't'
t=150/(125 cos θ)
0=-5[(150/125 cos θ)]^2 + 125 [150/(125 cos θ)] sin θ
This simplifies to
0=-5[1.44/(cos θ)^2]+150 tan θ
0=-7.2[1+(tan θ)^2]+150 tan θ
0=-7.2-7.2 (tan θ)^2+150 tan θ
Rearranging,
7.2 (tan θ)^2-150 tan θ+7.2=0
Using the Quadratics roots formula
tan θ=[-(-150) ± sqrt[22500-4(7.2)(7.2)]]/ 2 (7.2)
tan θ= (150 ±149.3072)/14.4
θ=87.25 degrees or 2.75 degrees
OR
θ=87°15' or 2°45' (converted into degrees and minutes as required in the solutions)