How do I solve this motion problem using the given equations?

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To determine the maximum height of a playground slide that allows a child to reach a speed of 15 m/s, the discussion highlights the use of energy conservation principles and kinematic equations. The key equation derived is h = (v^2)/(2g), where g is the acceleration due to gravity. By substituting the values, the calculated height is approximately 11 meters. The participant initially struggled with the provided equations but successfully applied energy concepts to solve the problem. The discussion emphasizes the importance of understanding both kinematic and energy equations in physics problems.
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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_{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\Deltay
therefore \Deltay = \frac{v_{y}^{2}}{-2g} = \frac{15^{2}}{-9.81\times-2}
=11m

Thanks
 
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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)
 
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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