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Three kinematics problems

  1. Jul 3, 2005 #1

    Zem

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    Air resistance is ignored in all three problems.
    1) A particle is launched at a 45-degree angle 1000m from the edge of a clff that is 500m high. It needs to land 1000m from the base of the cliff. What initial velocity is necessary?

    Y_f = y final = -500m
    x_f = x final = 2000m
    g = 9.81m/s^2

    I found this equation used for a similar problem, and was wondering if it works in this situation.
    y_f = (tan(theta)) * x_f - [g/2 * (v_i^2) * (cos^2(theta)] * x_f^2

    I got 125m/s. Is that right? Not sure if I'm using the right formula for this situation.

    2) A ski jump is designed so that the skier leaves the jump moving horizontally and then lands h = 10.0m below and D = 20.0m beyond the edge of the jump. Find (a) The time t the skier will remain in the air. (b) The speed v_i required to travel a distance D horizontally. (c) The vertical component of the skier's velocity upon landing. (d) The angle theta of the landing ramp so that the skier lands parallel to the ramp.
    v_yi = initial velocity in y direction
    v_yf = final y-velocity
    theta_landing_ramp = angle of the landing surface
    (a) Can I use t = sq_rt(2h/g) since v_yi = 0? I got 1.43s, which seems like a quick jump considering the skier went 20m.
    (b)v_xi = delta-x/delta-t = 14m/s?
    (c)v_yf = h/t = 7.00m/s ?
    (d)theta_landing_ramp = tan^-1(10m/20m) = 26.5 degrees?

    3) A ski-jumper leaves the ski track moving the horizontal direction with a speed of 25.0m/s. The landing incline below him falls off with a slope fo 35.0 degrees. Where does he land on the incline?
    v_yi = 25.0m/s
    x_f = horizontal distance from the edge of the track
    y_f = vertical distance
    a_y = -g
    d = hypotonuese
    (1)x_f = v_xi(t) = 25.0m/s * (t)
    (2)y_f = v_yi(t) + 1/2(a_y(t)^2) = -4.9m/s^2 * (t)^2

    (3) x_f = d * (cos(35.0))
    (4)y_f = -d * (sin(35.0))
    d * (cos(35.0)) = (25.0m/s)t
    -d * (sin(35.0)) = -4.9m/s^2(t)^2

    The book says: Solving (3) for t and substituting the result into (4), we find that d = 109m. Then they calculate x_f and y_f:
    x_f = d * (cos(35.0))
    Y_f = -d * (sin(35.0))

    But how did they solve for t and find d = 109m? (3) seems to have two unknowns in only one equation.

    All the Best,
    Zem
     
    Last edited: Jul 3, 2005
  2. jcsd
  3. Jul 3, 2005 #2
    you can use 'R = (v0^2 / g) * sin(2*theta)

    R = 2000
    g = 9.8
    then do the alg

    I got v = 140
     
  4. Jul 3, 2005 #3

    Zem

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    I was thinking the change in height had to be included.
    y_f = change in height = -500m
    -500m = (tan(45)) * 2000m - [9.8m/s^2/(2v_i^2) * (cos^2(45))] * 4x10^6m
    0m = (1) * 2500m - [9.8m/s^2/((2v_i^2) * (cos^2(45))] * 4x10^6m
    2500m = 4x10^6m * (19.6/2 * v_i^2)
    1600m * (19.6m/s^2) = 2 * v_i^2
    15,680m^2/s^2 = v_i^2
    v_i = 125m/s

    I saw that range formula in another thread, but thought it wouldn't work for this problem because there is a change in height. Am I right?

    Thanks!
     
  5. Jul 3, 2005 #4
    You are using the equation of a projectile for this one. If you use the equation R = (V0^2 / G) *sin(2 theta) you are taking into account the disp in Y.
     
  6. Jul 3, 2005 #5
    The answers to question number appers to be correct
     
  7. Jul 4, 2005 #6
    The way i did the first one was:

    the x_ vel. is v0*cos{@} 1

    the y_ vel is v0*sin{@} 2

    you can get the x_disp from v0*cos{@}*t 3

    and the Y_ disp from v0*sin{@}*t - 1/2 * g* t^2 4

    the time is derived from 3 x_disp / v0*cos{@} 5

    now you can use the relationship y_disp = x_disp * tan{&} 6

    now you can sub 5 and 6 into 4 and you get something like this after simplification

    tan{&} = tan {@} - 1/2 * (( g * x_disp ) / (v0^2 * cos{@} ^2)) 7

    and then

    x_disp = 2 * (v0 * cos{@}) ^2 * (tan{@} - tan{&}) / g) 8

    then you use the given values

    x_disp = 2000

    @ = 0

    & = 45

    g = -9.8

    and you come up with v0 = 98.99 m/s

    apparently my first post was in error my greatest apologies

    your solution for prob 2 seems to be okay

    as for three you can find the answer by the same method as above

    i get x_disp = 89.31 m

    then use r = x_disp / cos{&} = 109.03 m

    you can then plug this back into the eq. for x_disp or y_disp to get t = 3.57 s
     
  8. Jul 6, 2005 #7

    Zem

    User Avatar

    Thanks for your help, Mathmike!

    I asked about #1 in the help session with the professor, and it turns out my solution was right. :cool: Although he did say that equation only works when the trajectory starts from the origin, so in keeping with good form, he suggested doing the problem from scratch:

    h = 500m
    Yf = 0
    Xf = 2000m

    X = 0 Vi(cos(theta))(t)
    t = x/V(sin(theta))
    Yf = h + Vi(sin(theta))(t) - 1/2(gt^2)

    then:
    Yf = h + Vi(sin(theta))(x/Vi(cos(theta))) - 1/2(g(Xf^2/V^2(cos^2(theta))))

    Yf - h - Xf(tan(theta)) = -1/2(g(Xf^2/Vi^2(cos^2(theta))))

    (h + Xf(tan(theta) - Yf))(2(Vi^2(cos^2(theta)))) = gXf^2

    Vi^2 = (gXf^2/2(cos^2(theta))(h + Xf(tan(theta)) - Yf)

    Vi = (Xf/(cos(theta)))(sq_rt(g/(2(h + Xf(tan(theta))) - Yf)))
    = 2829(sq_rt(0.00196))
    = 125m/s
     
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