Projectile & Polynomial question (1 Viewer)

[lyounamu]

I have problems with Q6 b) ii & iii page 9

And I don't understand Q7 b) ii) & iii) page 10

Thanks in advance.

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_ext1_05.pdf

 

[foram]

6)

(i)
y= -4.9t^2 + 200t +5000
dy/dt = -9.8t + 200
at dy/dt = 0
9.8t = 200
t = 20.41 s
at t= 20.41 s
y= 7040.8 m

(ii)
when m= tan(-45)
t= 2x(20.41)
t= 40.42 s
at m = tan(-60)
m = sqrt.3
dy/dx = sqrt.3
dy/dx = dy/dt . dt/dx
dt/dx = 1/(dx/dt) = 1/200
dy/dt = -9.8t + 200
dy/dx = (-9.8t + 200)/200 = sqrt.3
9.8t/200 -1 = sqrt.3
t= 200(sqrt.3 +1)/9.8
=55.76 s

(iii)
dx/dt = 200
dy/dt = -9.8t +200
V = sqrt.[200^2 + (-9.8t +200)^2]

at V=350

350^2 - 200^2 = (-9.8t + 200)^2
9.8t = 200 + sqrt.82500
t = 49.72 s

 

[Managore]

7 b) ii)

Note that f(sqrt[3]/3) = A(sqrt[3]/9)-A(sqrt[3]/3)+1 = -A(2sqrt[3]/9)+1.

When A < 3sqrt[3]/2,

-A(2sqrt[3]/9)+1 > 0.

Also note that f(-sqrt[3]/3) = A(-sqrt[3]/9)-A(-sqrt[3]/3)+1 = A(2sqrt[3]/9)+1.

When A > 0,

A(2sqrt[3]/9)+1 > 0

Then you can draw a diagram or give an explanation similar to (but better worded than) "So since the local maximum and local minimum are both positive the curve can't cut the x-axis between its turning points, +/-sqrt[3]/3, and so will only cut the x-axis once".

iii)

Since A > 0, the curve is increasing in the region (-inf, -sqrt[3]/3) and since the curve is positive at f(-1) this means it has a zero for some x<-1. Since it only has one zero, it can't have a zero for -1<=x<=1.


I know this is badly worded but hopefully this gives you some idea.

 

[lyounamu]

6)

(i)
y= -4.9t^2 + 200t +5000
dy/dt = -9.8t + 200
at dy/dt = 0
9.8t = 200
t = 20.41 s
at t= 20.41 s
y= 7040.8 m

(ii)
when m= tan(-45)
t= 2x(20.41)
t= 40.42 s
at m = tan(-60)
m = sqrt.3
dy/dx = sqrt.3
dy/dx = dy/dt . dt/dx
dt/dx = 1/(dx/dt) = 1/200
dy/dt = -9.8t + 200
dy/dx = (-9.8t + 200)/200 = sqrt.3
9.8t/200 -1 = sqrt.3
t= 200(sqrt.3 +1)/9.8
=55.76 s

(iii)
dx/dt = 200
dy/dt = -9.8t +200
V = sqrt.[200^2 + (-9.8t +200)^2]

at V=350

350^2 - 200^2 = (-9.8t + 200)^2
9.8t = 200 + sqrt.82500
t = 49.72 s

I am sorry, Foram. I really don't understand how you worked out part ii and iii. I think in-depth explanation will do a job. Sorry!

 

[lolokay]

ii) v_y must be between -200 and -200 sqrt[3]
and v_y = 200 - 9.8t

hopefully you can see why that is


iii) 350 is the vector sum of the x and y components of velocity
v = 350 when v_y² = 350^2 - 200^2
and v_y = 200 - 9.8t