Applied Maths Exam Question

alan4cult

Hello,

Anybody done Questions on the 1986 Applied Maths Paper? I am looking for help on Question 3, the projectiles question. For anybody who doesn't have the question here it is:

A particle is projected with speed u at an angle α to the horizontal. If the maximum height reached is the same as the total horizontal range, show that tan α = 4.

The particle is moving at right angles to its original direction of motion after a time t1 and then strikes the horizontal plane after 8 seconds, both times measured from the instant of projection.

Show u =g root17
Calculate t1

I'm able to to every part except calculate t1. Any ideas?

JC 2K3

If it's at right angles to it's original direction of motion it means its velocity at that point will be equal to -1/u(product of slopes of 2 perpendicular lines is -1) which in this case equals -1/g(root17).

Now resolve the j componant of the velocity of the particle at that point and use v=u+at to get the time.

so:
uj = g(root17)sinA
vj = -sinA/g(root17)
aj = -g

-sinA/g(root17) = g(root17)sinA - gt1
t1 = (-sinA/g(root17) - g(root17)sinA)/-g)
t1 = root17(sinA) + sinA/(root17)
sinA = 4/root17
t1 = 4 + 4/17
t1 = 72/17


I think that's right, not 100% sure though.

EDIT: Forgot the minus sign for vj, thanks Zorba.

ZorbaTehZ

Realise that at that time the velocity is towards the ground i.e. negative.

That changes the end result to

t1 = 4 + 4/17
t1 = 4.24 seconds

Very nice though JC2K3.

alan4cult

nice one

JC 2K3

Ah, forgot the minus. Post edited, thanks Zorba.

alan4cult

I'm confused as to how you are resolving the j component. Take a look at my diagram. Should it be -cosa/groot17?

ZorbaTehZ

You're diagram is wrong mate.
Consider the motion described by this particle (in your head)
Its line of movement is perfectly symmetrical, since gravity is the only force effecting it (obviously we disregard air resistance etc)
What I mean by this is that since it is symmetrical its take off angle is the same as its landing angle i.e. stick the alpha on the groundline.

EDIT: hmm Im looking at the diagram again and im not soo sure now what it is off? *confused* possibly scratch what I just said

alan4cult

ZorbaTehZ wrote:

You're diagram is wrong mate.
Consider the motion described by this particle (in your head)
Its line of movement is perfectly symmetrical, since gravity is the only force effecting it (obviously we disregard air resistance etc)
What I mean by this is that since it is symmetrical its take off angle is the same as its landing angle i.e. stick the alpha on the groudline.

I get you now. Its much simpler your way. Your reading the diagram from right to left as if the intial velocity was the final velocity and resolving away from the horizontal. I gotta stop over complicating things.

Cheers for all the help guys.

alan4cult

ZorbaTehZ wrote:

EDIT: hmm Im looking at the diagram again and im not soo sure now what it is off? *confused* possibly scratch what I just said

Always go with instinct!

ron-burgandy

what's the solution to the first part "show that tan α = 4."?

I really need to start studying this subject

JC 2K3

s perpendicular max = s parallel total

at max height, t = T/2 (T= total time of flight)

perp:
s max = u(sinA)T/2 - g(T^2)/8

parallel:
s total = u(cosA)T

8u(cosA)T = 4u(sinA)T - g(T^2)
divide by TcosA:
8u = 4u(tanA) - gT/(cosA)
4u(2 - tanA) + gT/(cosA) = 0
2 + gT/4u(cosA) = tanA

Now, to find T:
perp:
0 = u(sinA)T - g(T^2)/2
u(sinA) = gT/2
2u(sinA) = gT

2 + 2u(sinA)/4u(cosA) = tanA
2 + tanA/2 = tanA
4 + tanA = 2tanA
tanA = 4

1337 Differentiation Method:
If you draw the diagram of the curve and mark the max height and total distance as 2k you'll notice that you have 3 points:
starting point = (0,0)
max height = (k,2k)
end point = (2k,0)

Since the slope of a line is equal to the tan of the angle it makes with the horizontal we need to get dy/dx at (0,0)

curve of particle = ax^2 + bx + c
dy/dx = 2ax + b

at starting point:
curve = a(0) + b(0) + c = 0
c = 0 ................(1)

dy/dx = 2a(0) + b = b
b = tanA ........................(4)

at max height:
ak^2 + bk + c = 2k .......................(2)

dy/dx = 2ak + b = 0
a = -b/2k .......................(5)

at end point:
4ak^2 + 2kbx + c = 0 ..................(3)

dy/dx = 4ak + b .......................(6)



ak^2 + bk + c = 2k .....................(2)
c = 0 ........................(1)
ak^2 + bk = 2k
ak + b = 2
a = -b/2k .......................(5)
-b/2 + b = 2
-b + 2b = 4
b = 4
b = tanA ........................(4)
tanA = 4

Shox

how do you even get papers back this far?

alan4cult

Shox wrote:

how do you even get papers back this far?

If you go to www.discoveringmaths.com and to the applied maths section of the website you will get an e-mail address for oliver murphy. If you contact him he will send you a booklet of exam papers 1983 - 2006. A booklet costs about 15-20 euro.