Question

problem 6:
a water slide is 72 ft high. how fast do you go when you reach the bottom of
the slide (ignoring friction effects). how fast do you go at the halfway point?
does the shape of the slide matter?

Explanation:

Step1: Apply conservation of mechanical energy

The potential energy at the top \(U = mgh\) (where \(m\) is mass, \(g = 32\ ft/s^{2}\) is acceleration due to gravity, \(h\) is height) is converted to kinetic energy \(K=\frac{1}{2}mv^{2}\) at the bottom. By conservation of energy \(mgh=\frac{1}{2}mv^{2}\), so \(v = \sqrt{2gh}\).
For \(h = 72\ ft\), \(v=\sqrt{2\times32\times72}\)

$$v=\sqrt{4608}=67.9\ ft/s$$

Step2: For the halfway point \(h'= 36\ ft\)

Using \(v'=\sqrt{2gh'}\), \(v'=\sqrt{2\times32\times36}\)

$$v'=\sqrt{2304} = 48\ ft/s$$

Step3: Analyze the effect of slide - shape

Since we use conservation of mechanical energy (\(mgh=\frac{1}{2}mv^{2}\)), which only depends on the vertical height change (because \(h\) in \(mgh\) is the vertical height). The shape of the slide (as long as friction is ignored) does not matter.

Answer:

• At the bottom: \(67.9\ ft/s\)
• At the halfway point: \(48\ ft/s\)
• The shape of the slide does not matter.