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Mechanics -projectile motion calculating angle

  1. May 19, 2012 #1
    1. The problem statement, all variables and given/known data
    A stone is thrown from a building of height 40m and reaches a maximum height of 49m from the ground. The stone strikes the ground at a distance of 10m from the foot of the building. find the angle of elevation from which the stone is projected. Assuming the ground level is horizontal.


    2. Relevant equations
    Hmax=Vo2sin2θ/2g

    3. The attempt at a solution
    from the hmax formula i found Vosinθ=13.28m/s
     
    Last edited: May 19, 2012
  2. jcsd
  3. May 19, 2012 #2
    y=uyt- 0.5gt2
    x=uxt
     
    Last edited: May 19, 2012
  4. May 19, 2012 #3
    That is incorrect. Hmax=h + Vo2sin2θ/2g is correct. The equation you've written gives the range of the projectile(not the maximum height.) :smile:


    To the OP,
    What is the horizontal distance traveled by the stone in terms of velocity and t? And the vertical distance traveled by the stone in terms of t?
     
  5. May 19, 2012 #4
    from y=uyt- 0.5gt2
    t=4.8s
    if substituted in x=uxt
    Ux=2.08s=Vo
    hence sinθ=6.37 which is invalid
     
    Last edited: May 19, 2012
  6. May 19, 2012 #5
    Sorry about that, assuming 'H' for horizontal and and I was wrong assuming vertical displacement=0, which not in this case.
     
  7. May 19, 2012 #6
    How do you find t when there are 2 unknown variables, t and uy. in 1 equation?
     
  8. May 19, 2012 #7
    Uy is Usinθ=13.28m/s
    which i found from Hmax equation:
    Usinθ=√9×2×9.8=13.28
     
  9. May 19, 2012 #8
    The equation Hmax=Vo2sin2θ/2g applies to when vertical displacement equal to zero.
    In this case the stone does not drop to the same level as it was projected.
     
  10. May 19, 2012 #9

    yeah and that is why is did not use height 49 but i used height 9m...hope i'm understanding what you meant :S
     
  11. May 19, 2012 #10
    Ok it is correct if you substitute 9m for the equation.
    What you get is the time taken from start to the same level(on descending) of the flight.
    But there another 40m to travel down. Now you are making 2 segments of flight.
    It is best to remember simple formula for displacement of constant acceleration.

    Δs=s0+ut ±.5at2

    If at t=0 and s0[/SUB=0
    Δs=ut ±.5at2

    If at t=0 , s0=0 and a=0
    Δs=ut

    From these 2 equations you can derive all other equations depends on Δs.
     
  12. May 19, 2012 #11

    rl.bhat

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    Use -y = Usinθ*t - 1/2*g*t^2. Substitute the values for y, Usinθ and g and solve for t.
     
  13. May 19, 2012 #12
    here this is how i get...did i substitute wrong?
     
  14. May 19, 2012 #13
    I am new to this thing and dunno how to post my own question so thats the first thing. Second my problem is I know displacement is 20m, acceleration is -5.4m/s, final velocity is 0m/s, coefficient of friction=0.55, the object slowing down weighs 1000kg, can i find initial velocity and the time it took to slow down and if so how?
     
  15. May 19, 2012 #14

    rl.bhat

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    -40 = 13.28*t - 4.9*t^2
    Now solve for t.
     
  16. May 19, 2012 #15
    t=4.8s
    UCosθt=10
    UCosθ=10/4.8

    USinθ/UCosθ=13.28 x 4.9/10
    θ=81°
     
  17. May 19, 2012 #16
    thanks :D
     
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