Projectile Motion Question (1 Viewer)

[Trebla]

Hi guys, can you help me with this question? It's a past HSC Question (1984) and can also be found in the syllabus:

Two stones are thrown simultaneously from the same point in the same direction and with the same non-zero angle of projection (upwards inclination to the horizontal), α, but wiuth different velocities U, V metres per second (U<V).

The slower stone hits the ground at a point P on the same level as the point of projection. At that instant the faster stone just clears a wall of height h metres above the level of projection and its (downward) path makes an angle β with the horizontal.

a) Show that, while both stones are in flight, the line joining them has an inclination to the horizontal which is independent of time. Hence, express the horizontal distance from P to the foot of the wall in terms of h and α.

b) Show that:

V(tan α + tan β) = 2U tan α

and deduce that, if β = α/2, then

U < 3V/4



Thanks...

 

[Trebla]

Um....help...............*echoes*

 

[_ShiFTy_]

I got hcotα for part a...

 

[Trebla]

How did you get that?

 

[_ShiFTy_]

I'm not sure if thats the answer but heres what i did

For the slower ball (#1)

x = utcosα
y = -gt²/2 + utsinα

y = 0 to find the time when it lands and you would get
t = 2usinα/gWhen t = 2usinα/g
x = 2sinαcosα.u²/g
x = sin2α.u²/g......(1)
--------------------------------------------
For the faster ball (#2)
x = vtcosα
y = -gt²/2 + vtsinα

When t = 2usinα/g, y = h
x = sin2α.uv/g......(2)
h = u(v - u)/g x 2sin²α

The horizontal distance from P to the foot of the wall would be 'x' from ball #2 minus 'x' from ball #1
i.e. (2) - (1)
D = usin2α(v - u)/g.....(3)

But h = u(v - u)/g x 2sin²α
u(v - u)/g = h/2sin²α .....(4)
Sub (4) into (3) which gives you

D = sin2α . h/2sin²α
= 2sinαcosα . h/sin²α
D = hcotα


There is probably a much simplier way of doing his but thats what i did..

 

[Riviet]

I got hcotα as well for the distance:

At time t,

v(vtcosα , -gt²/2 + vtsinα) and u(utcosα , -gt²/2 + utsinα) where v and u represent the points where the two stones are with initial velocity V and U.

Now the inclination of the line joining V and U is given by m=(y2-y1)/(x2-x1) which gives tanα and is clearly independent of t.

If you draw a diagram and label P, some point on the path of the faster stone (V) and the foot of this point, join the 3 points to form a right triangle, we can deduce that the inclination is given by:

m=tanα=h/D

.'. D=hcotα

 

[_ShiFTy_]

eep, forgot to read the 1st part of the question...so my way is wrong since it says hence

 

[Yip]

tan β=(dy/dt)/-(dx/dt) [dx/dt is negative since u are observing the projectile from the other end hence x will be decreasing]
=[-g(2usinα/g)+vsinα)/-vcosα]
=-tanα+(2u/v)tanα
v(tanα+tan β)=2utanα

If β = α/2,
v[tanα+tan(α/2)]=2utanα
Let tan(α/2)=t,
v[t+(2t/1-t^2)]=2u(2t/1-t^2)
2vt+vt(1-t^2)=4ut
t^2=(3v-4u)/v>0
u<3v/4

 

[Roobs]

Hmm...i've just hit this question too and its got me confused--- I keep coming up with

2UTan a = V( tan a- tan b) and not the " + " that should be there-- the part i dont get, (and maybe you can help me out with a more detailed explanation yip) is why dx/dt is negative?

The worked solutions i have to this problem use the components of velocity (y* and x*) to find the gradient, using "Tan b = -y*/x*, and there i hit the wall again--- isnt gradient given by m= tan x?

Im stuck thinking that as y* is directed down (ie negative) using -y*/x* will yeild a positive gradient which clearly isnt correct??? can anyone clarify?

the other way i attempted this, was by using the cartesian equation of motion, and its derivative, as the gradient function

where tan b = f'(x) ~~ where f(x) is cartesian equation ~~

but again i cant figue out why/ where i need to "throw a negative in" to get the requred result

can someone spell this out for me?

 

[Roobs]

And another thing-- ive just done another simmilar question ((fitzpatrick 35(f) p186 q15)

"A ball thrown from a point A with speed V, at an inclination a to the horizontal reaches a point B after t seconds. Find the position of B relative to A and show that if AB is inclined at <theta> to the horizontal then the direction of motion of the ball wehn at b, is inclined to the horizontal at an angle <phi> given by

tan <phi> = 2tan <theta> - tan a

here i used the cartesian equation again, found the derivative, which gave gradient, ie tan <phi>, but used no negative gradient et, and it just fell out.....this aparent inconsistency is shitting me......

 

[Yip]

why dx/dt is negative?

dx/dt is negative because since the downward path makes an angle of beta with the horizontal (acute ange I presume), its easiest to redefine x as the distance from the point of impact. Hence, as the particle approaches the point of impact, the value of x decreases and hence dx/dt is negative. note that dy/dt is not negative, as regardless of whether x is defined as the distance from point of impact or distance to point of impact, it will be the same. at least, thats how i interpret it...

 

[haque]

Rather than taking dx/dt as negative, we can see that the angle b with the horizontal is made on the particle's descent and the angle b is measure clockwise from the horizontal rather than anticlockwise(or u coiuld say that b lies in the second qudrant in which we get -tanb rather than tanb for dy/dt/dx/dt. hence the result holds.

 

[Roobs]

hmm..im not quite getting both of the posts above me...i think it's time for pictures...i scanned my working, and if someone could have a look and tell me/put a big ass red circle around where i've gone wrong it would be appreciated, thanks...... (had to put them in a word doc. for the filesize limit)....


Hmm...i think i might have solved my own problem with my second diagram there,have a look people, and tell me if this is just an issue of how the angle b is defined.....

 

[haque]

the angle u had marked beta is in actual fact 180-beta and tan(180-beta) is -tanbeta. which is what i was trying to say in my previous post.

 

[Roobs]

hmm..ok.. i think im getting it now-- so by convention does "and angle x with the horizontal" refer to the acute angle, or the angle made with the positive direction of the x axis?...could someone do the fitzpatrick q15 (quoted above) by the same principals, and tell me if they run into any problems.....

Thanks everyone for their input

 

[haque]

by convention it's the acute angle.