Question

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

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

Hint:

In swing problems, use conservation of energy: loss of potential energy \(mgh\) becomes kinetic energy \(\frac{1}{2}mv^2\).

Answer 1

The Correct Option is C

At the maximum height, the girl is: \[ 3\text{ m} \] above the ground. At the lowest point, the girl is: \[ 2\text{ m} \] above the ground. So, the vertical drop is: \[ h=3-2. \] \[ h=1\text{ m}. \] At the maximum height, velocity is momentarily zero, so the energy is mainly gravitational potential energy. At the lowest point, this loss of potential energy becomes kinetic energy. Using conservation of mechanical energy: \[ mgh=\frac{1}{2}mv^2. \] Cancel \(m\) from both sides: \[ gh=\frac{1}{2}v^2. \] Therefore: \[ v^2=2gh. \] Substitute: \[ g=9.8\text{ m/s}^2,\qquad h=1\text{ m}. \] \[ v^2=2(9.8)(1). \] \[ v^2=19.6. \] Taking square root: \[ v=\sqrt{19.6}\text{ m/s}. \] Hence, the maximum velocity is: \[ \sqrt{19.6}\text{ m/s}. \]