Motion equations problem

  • Thread starter faoltaem
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  • #1
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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
 

Answers and Replies

  • #2
3
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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)
 
  • #3
31
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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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