# Motion problem

1. Sep 13, 2007

### faoltaem

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$$_{x}$$ = v$$_{0x}$$ + a$$_{x}$$t
x = $$\frac{1}{2}$$ (v$$_{0x}$$ + v$$_{x}$$)t
x = v$$_{0x}$$t + $$\frac{1}{2}$$a$$_{x}$$t$$^{2}$$
v$$_{x}$$$$^{2}$$ = v$$_{0x}$$$$^{2}$$ + 2a$$_{x}$$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$$_{max}$$ = 15m/s where g = -9.81m/s
v$$_{y}$$$$^{2}$$ = -2g$$\Delta$$y
therefore $$\Delta$$y = $$\frac{v_{y}^{2}}{-2g}$$ = $$\frac{15^{2}}{-9.81\times-2}$$
=11m

Thanks

2. Sep 13, 2007

### proasd

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. Sep 13, 2007

### faoltaem

thanks for that, i didn't even consider energy and those formulae are included too so no memorising

once again thanks a bunch