Question

1. a 60 - kg swimmer at a water park enters a pool using a 2 m high slide.

a) find the velocity of the swimmer at the bottom of the slide

Explanation:

Step1: Identify the physics principle (Conservation of Mechanical Energy)

Assume no friction, so initial gravitational potential energy (\(PE\)) is converted to final kinetic energy (\(KE\)).
\(PE = mgh\), \(KE=\frac{1}{2}mv^2\)
Set \(PE_{initial}=KE_{final}\) (since initial \(KE = 0\) at rest, final \(PE = 0\) at bottom).

Step2: Set up the energy conservation equation

\(mgh=\frac{1}{2}mv^2\)
Mass \(m\) cancels out: \(gh=\frac{1}{2}v^2\)

Step3: Solve for velocity \(v\)

Rearrange: \(v = \sqrt{2gh}\)
Given \(g = 9.8\,\text{m/s}^2\), \(h = 2\,\text{m}\)
\(v=\sqrt{2\times9.8\times2}\)
Calculate: \(2\times9.8\times2 = 39.2\), so \(v=\sqrt{39.2}\approx6.26\,\text{m/s}\)

Answer:

The velocity of the swimmer at the bottom of the slide is approximately \(\boldsymbol{6.3\,\text{m/s}}\) (or more precisely \(\sqrt{39.2}\approx6.26\,\text{m/s}\)).