Question

At her maximum height a girl in a swing is \(3m\) above the ground and at the lowest point she is \(2m\) above the ground. Her maximum velocity is

• A. \(\sqrt{29.4}\,m/s\)
• B. \(\sqrt{9.8}\,m/s\)
• C. \(\sqrt{19.6}\,m/s\)
• D. \(9.8\,m/s\)

Hint:

Use conservation of energy: loss of potential energy equals gain in kinetic energy.

Answer 1

The Correct Option is C

Concept:
In a swing, potential energy at maximum height converts into kinetic energy at the lowest point.

Step 1: Maximum height above ground: \[ h_1=3m \]

Step 2: Lowest height above ground: \[ h_2=2m \]

Step 3: Fall in height: \[ h=h_1-h_2=3-2=1m \]

Step 4: By conservation of energy: \[ mgh=\frac{1}{2}mv^2 \]

Step 5: Cancel \(m\): \[ gh=\frac{v^2}{2} \] \[ v^2=2gh \]

Step 6: Put \(g=9.8\,m/s^2\), \(h=1m\): \[ v=\sqrt{2(9.8)(1)} \] \[ v=\sqrt{19.6}\,m/s \] \[ \boxed{\sqrt{19.6}\,m/s} \]