Turn on thread page Beta

projectile motion questions watch

Announcements
    • Thread Starter
    Offline

    1
    ReputationRep:
    Fairly easy, however I have forgotten all this stuff

    A coastal defence cannon, positioned at the top of a 150m vertical cliff, is able to fire shells at a velocity of 120ms. If fired horizontally out from the cliff, how far out to sea will the shell land?

    (b) The cannon is now aimed up at an angle of 45 degrees above horizontal. How far will a shell now travel?
    Offline

    19
    ReputationRep:
    (a)vertically,
    u= 0 a= 9.8 x= 150

    x= ut + 0.5at^2

    work out t then subs horizontally into x= ut where u=120 to find x.

    (b) vertically, u is now u= 120sin45 and horizontally u = 120cos 45 (they're actually the same). use same method, find t then horizontal distance.
    Offline

    0
    ReputationRep:
    a=-9.8
    s=-150
    u=0
    t= ?
    s = ut +1/2 at^2
    -150 = -4.9t^2
    t=5.5s
    s= ?
    u = 120
    a = 0
    s = ut + 0.5at^2
    s = 663m

    Same again except
    u = 120sin45 for first part
    and u = 120cos45 for second
    • Thread Starter
    Offline

    1
    ReputationRep:
    ok im doing the first part for part b, using your methods. You get a quadratic, however when you try to solve you get two values for t. Which value do you use?
    Offline

    19
    ReputationRep:
    Sorry should have been clearer on the second one.

    If you take up as positive, then vertically
    u= 120sin45 a=-9.8 (opposing) x= -150
    (sin45= 1/√2 )

    -150= 120t/√2 - 4.9t²
    4.9t² -120t/√2 -150=0

    If you solve that you get one negative and on positive value, obviously t can only be positive.
    • Thread Starter
    Offline

    1
    ReputationRep:
    I get 137.2m, does that sound right?
    Offline

    19
    ReputationRep:
    nope. think about it, if you fire it up at an angle, it will travel even further than before.

    solving 4.9t² -120t/√2 -150=0
    i get t= 18.93s

    then distance horizontally,
    x= 18.93*120/√2
    x= 1607m
    • Thread Starter
    Offline

    1
    ReputationRep:
    Yeah I was thinking about that. Thanks for the help, I missed a - in the quadratic formula which probably threw it off (dont do physics at midnight).
    • Thread Starter
    Offline

    1
    ReputationRep:
    What about this question?

    A rifle is fired up at an angle of 55 degrees above horizontal. If the initial velocity of the bullet is 570ms, what wil lbe its velocity (magnitude and direction) 2.5 s after firing?

    x component = 570cos55
    y component = 570sin 55

    So I guess we just find delta y, and then use that in the equation (v^2 = u^2 + 2a(delta y)) ?
    Offline

    0
    ReputationRep:
    (Original post by bally)
    What about this question?

    A rifle is fired up at an angle of 55 degrees above horizontal. If the initial velocity of the bullet is 570ms, what wil lbe its velocity (magnitude and direction) 2.5 s after firing?

    x component = 570cos55
    y component = 570sin 55

    So I guess we just find delta y, and then use that in the equation (v^2 = u^2 + 2a(delta y)) ?
    It's been a while since A'level ...

    v = u + at

    .: Vy = (570sin55) + (-9.8)(2.5)
    .: Vy = 442.4166652 ms^-1

    => V = sqrt[(442.4166652)^2 + (570cos55)^2]
    .: V = 550.1 ms^-1

    => theta = arctan[442.4166652 / (570cos55)]
    .: theta = 53.5


    So, velocity is 550.1 ms-1 at 53.5o to the horizontal.
    Offline

    0
    ReputationRep:
    (Original post by bally)
    thanks, but i dont get where this equation comes from?

    => V = sqrt[(442.4166652)^2 + (570cos55)^2]
    It's just Pythagoras. You have the new y component and the x component and need to find the resultant...
    • Thread Starter
    Offline

    1
    ReputationRep:
    edit; lol just realised in time (about pythagoras)

    => theta = arctan[442.4166652 / (570cos55)]
    .: theta = 53.5

    Isnt the 442.... the hypotenuse and the 570cos55 the bottom of the triangle? How does that work?

    edit: oh wait this is the new triangle where we resolve the forces to find the resultant angle...
    Offline

    0
    ReputationRep:
    (Original post by bally)
    edit; lol just realised in time (about pythagoras)

    => theta = arctan[442.4166652 / (570cos55)]
    .: theta = 53.5

    Isnt the 442.... the hypotenuse and the 570cos55 the bottom of the triangle? How does that work?
    The 442 is the vertical (side) and 570cos55 is the horizontal (bottom).

    We know that: tan theta = opposite / adjacent

    and so therefore theta = arctan[opposite/adjacent]. Arctan is just the inverse tan function, sometime written as tan-1
    Offline

    0
    ReputationRep:
    (Original post by bally)
    ed
    edit: oh wait this is the new triangle where we resolve the forces to find the resultant angle...
    yep
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: July 9, 2006

University open days

  • University of Lincoln
    Brayford Campus Undergraduate
    Wed, 12 Dec '18
  • Bournemouth University
    Midwifery Open Day at Portsmouth Campus Undergraduate
    Wed, 12 Dec '18
  • Buckinghamshire New University
    All undergraduate Undergraduate
    Wed, 12 Dec '18
Poll
Do you like exams?

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.