Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta

Mechanics question - Maximising horizontal distance watch

    • Thread Starter
    Offline

    0
    ReputationRep:
    Any advice woud be much appreciated.

    The question states: 'A stone is thrown at an angle theta to the horizontol from a cliff 100m above the sea at 30m/s. At what angle should it be thrown to maximise the horizontol distance it travels?'

    Now i understand what you have to do, but the algebra gets ridiculous...

    x = Vtcosθ i + (Vtsinθ - 1/2gt²) j (where x = displacement, V = start speed, θ = angle thrown, g = acceleration due to gravity = 9.8 , t = time after throw , i and j are vectors)

    Therefore, using just the 'j' component of displacement,

    -100 = 30tsinθ - 1/2gt² (minus 100 because the stone starts 100m above it)

    and solving for t using quadratic formula, gives

    t = 10/g (3sinθ + (9sin²θ + 2g)^½)

    Now we use the displacement vector for just i

    w = 30cosθ t (from first equation), where w = horizontol distance.

    => w = 30cosθ (10/g (3sinθ + (9sin²θ + 2g)^½)) (subbing for t)

    now we have to differentiate this to find the maximum, and set the differentiated bit to 0, and solve for θ. YEEESSSSSHHHH..i've done the differentiation and i've tried so many different ways of solving for θ. Can anyone help, or can anyone suggest an alternative?

    In case anyone out there can be bothered to try this, after the differentiation you get:

    3cos2θ - sinθ(9sin²θ + 2g)^½ + 9cos²θsinθ(9sin²θ + 2g)^-½ = 0

    If anyone can solve that for theta, i'll buy them a pint...

    Cheers
    Offline

    1
    ReputationRep:
    i think there is a standard result for the maximum horizontal distance and the answer is 45 degrees. i'm not sure how to prove it but i'll try anyway.

    first we find what the time of flight is. let this be T.
    vertical components: speed(u)=Usin(alpha), acceleration(a)=-g, displacement(S)=0, time(t)=T
    using the eq'n S=ut + 1/2at^2,
    Usin(alpha)T - 1/2gT^2=0
    factorising, T=0 or T=2Usin(alpha)/g

    now for the horizontal distance:
    horizontal components: u=Ucos(alpha), S=R, t=T
    using the eq'n S=ut,
    R= Ucos(alpha)T
    = 2U^2sin(alpha)cos(alpha)/g
    simplying using the double angle formula,
    R= U^2sin2(alpha)/g

    for the horizontal distance to be maximum, sin2(alpha) must be maximum.
    sin2(alpha)=1
    therefore 2(alpha)=90
    and alpha=45

    and this is a standard result.
    this looks very messy especially since i don't know how to use symbols to indicate angles..sorry!
    basically how i did it was:
    a) find time of entire journey
    b) find horizontal range and substitute expression in a) for time.
    c) find angle for range to be maximum.

    hope this helps!
    Offline

    3
    ReputationRep:
    That is only true when the projectile stops at the same level from where it is thrown, here it is stopping 100m below where it starts from.
    Offline

    3
    ReputationRep:
    I've been trying this one for ages, and have to admit I am stumped. I can solve it using programs like mathematica. But doing it analytically is doing my head in. I get a quartic in theta squared. Which is just nasty. I guess I am missing some kind of trigonometric substitution along the line somewhere which greatly simplifies things .
    Offline

    12
    ReputationRep:
    Looks nasty! :eek:

    Looks like a lot of manipulation is needed! :eek:

    Don't look at me!

    FERMAT!!!
    Offline

    0
    ReputationRep:
    [u = initial speed, θ = required angle, x = horizontal distance, y = vertical distance, t = time]

    In order for the particle to reach maximum x, it will need to be in the air for the maximum amount of time.

    y = utsinθ - 0.5gt²

    Now, we need to obtain the time the particle strikes the ground. That is, the first time y=0:

    0 = utsinθ - 0.5gt²
    t(usinθ - 0.5gt) = 0
    t=0 or t=2usinθ/g (maximum t)

    Then:
    x = utcosθ = (2u²sinθ/g) * cosθ = u²2sinθcosθ/g, since x is maximum when t is maximum.
    But, 2sinθcosθ = sin2θ
    x = u²sin2θ/g

    x = 900sin2θ/g

    Since -1<sin2θ<1, x is max when sin2θ=1:
    sin2θ = g/900
    2θ = 0.62
    θ = 0.31

    I guess I messed up somewhere...
    Offline

    3
    ReputationRep:
    Yes, you forgot that the particle starts 100m above the "ground". And it simply isn't true that the particle has to be in the air for the longest time to to travel the furthest. If you directed it vertically upwards then it would be in the air for the longest time. But not travel horizontally at all.
    Offline

    0
    ReputationRep:
    I think I found my mistake. I overlooked the fact that the stone was 100m above sea level. So, the first time the ball strikes the ground isn't when y=0, it's when y=-100. That is:

    100 = utsinθ - 0.5gt²

    Solve that to get two values for t. Use both values in v=utcosθ, and continue normally.

    I hope I got it right this time.
    Offline

    0
    ReputationRep:
    (Original post by AntiMagicMan)
    And it simply isn't true that the particle has to be in the air for the longest time to to travel the furthest. If you directed it vertically upwards then it would be in the air for the longest time. But not travel horizontally at all.
    True, but in this case I think what I did applies?
    Offline

    3
    ReputationRep:
    If you find the value of t in terms of theta, put that into the equation for x then you still have the problem of maximising x.

    Finding x in terms of theta is easy, it is then differentiating that expression and maximising x which is the hard part.
    Offline

    0
    ReputationRep:
    (Original post by AntiMagicMan)
    If you find the value of t in terms of theta, put that into the equation for x then you still have the problem of maximising x.

    Finding x in terms of theta is easy, it is then differentiating that expression and maximising x which is the hard part.
    Yes, it does get quite complicated.
    Offline

    2
    ReputationRep:
    GOTTIT!!

    sin²ø = V²/[2(V² - gs)]

    Where,
    ø = angle of projection,
    V = speed of projection,
    s = distance below hill.

    Just geting my claim in.
    Writing it up now
    Offline

    3
    ReputationRep:
    (Original post by Fermat)
    GOTTIT!!

    sin²ø = V²/[2(V² - gs)]

    Where,
    ø = angle of projection,
    V = speed of projection,
    s = distance below hill.

    Just geting my claim in.
    Writing it up now
    That doesn't make any sense. V^2 is 900, g*s is 981, so you are saying that Sin(theta) is complex.

    By brute force I think that the answer is theta = 0.511076 and x = 163.6m. But that is using maple. Maybe that was a mistake by Fermat, I would be very interested in a proper proof though.
    Offline

    2
    ReputationRep:
    As some have noted, the working out/algebra was like to be pretty horrible, and it would be difficult to follow the working if it wasn't formatted properly.
    However, it's been ages since I did any latex, so I just wrote it out and scanned it in. Hope you can read it/follow it.

    I started at the eqns for the time of flight, and ended with an expression for maximising the (horizontal) distance travelled, depending on the angle of trajectory.

    Oh, yes. In the formula,

    sin²ø = V²/[2(V² - gs)]

    upwards is measured as +ve, so s should be entered as s = -100, for example.

    Edit: latex'd the proof.
    Attached Images
  1. File Type: pdf Projectile.pdf (48.0 KB, 125 views)
    Offline

    3
    ReputationRep:
    Nice. Very impressive. And that gives the same theta I obtained using maple.
    Offline

    2
    ReputationRep:
    (Original post by AntiMagicMan)
    Nice. Very impressive. And that gives the same theta I obtained using maple.
    Ah, great. Now I feel my work is justified.
    I'm afraid my mathematica has run its course though
    Now I'm using sci-lab
    Offline

    3
    ReputationRep:
    Oooh nice, Open Source.

    I gave up trying to solve for theta once I had got to an equation that had the sixth power of Sine in it :s. I haven't read your solution thoroughly but given that it gives the same answer as Maple, it is almost certainly right. I will when I get a bit of time. I guess I just need to brush up on Trig formulas.
    Offline

    12
    ReputationRep:
    More rep for Fermat I feels
    Offline

    2
    ReputationRep:
    (Original post by shiny)
    More rep for Fermat I feels
    • Thread Starter
    Offline

    0
    ReputationRep:
    Well i said i'd buy you a pint, Fermat....thanks for doing that.
 
 
 
Turn on thread page Beta
Updated: September 2, 2004
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.