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Motion problem

  1. Sep 13, 2007 #1
    I have a problem from a practice physics exam:
    The specifications for design of a playgroud slide say that a child should gain a speed of no more than 15m/s by sliding down the eqipment. How tall can the slide be?

    these are the equations that we'll be given in our exam:
    v[tex]_{x}[/tex] = v[tex]_{0x}[/tex] + a[tex]_{x}[/tex]t
    x = [tex]\frac{1}{2}[/tex] (v[tex]_{0x}[/tex] + v[tex]_{x}[/tex])t
    x = v[tex]_{0x}[/tex]t + [tex]\frac{1}{2}[/tex]a[tex]_{x}[/tex]t[tex]^{2}[/tex]
    v[tex]_{x}[/tex][tex]^{2}[/tex] = v[tex]_{0x}[/tex][tex]^{2}[/tex] + 2a[tex]_{x}[/tex]x

    although i can work out the answer to the problem i had to look up a new equation because i couldn't work out how to do it with the equations that we were given, so i was wondering if it was possible to do it with one of them, so if you could tell me which one would work the best or if i should just start memorising some new equations, but here's what i did anyway:
    v[tex]_{max}[/tex] = 15m/s where g = -9.81m/s
    v[tex]_{y}[/tex][tex]^{2}[/tex] = -2g[tex]\Delta[/tex]y
    therefore [tex]\Delta[/tex]y = [tex]\frac{v_{y}^{2}}{-2g}[/tex] = [tex]\frac{15^{2}}{-9.81\times-2}[/tex]
    =11m

    Thanks
     
  2. jcsd
  3. Sep 13, 2007 #2
    I would work out the energy
    gain in kinetic energy = loss in gravitational potential energy
    (1/2)(m)(v^2) = (m)(g)(h)
    simplifying you get
    v^2 = 2(g)h
    h = (121)/(2g)
     
  4. Sep 13, 2007 #3
    thanks for that, i didn't even consider energy and those formulae are included too so no memorising

    once again thanks a bunch
     
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