Question

If a body is dropped freely from a height of 20 m and reaches the surface of a planet with a velocity of 31.4 m/s, then the length of a simple pendulum that ticks seconds on the planet is:

• A. \(1 \, \text{m}\)
• B. \(0.625 \, \text{m}\)
• C. \(2.5 \, \text{m}\)
• D. \(2 \, \text{m}\) \bigskip

Hint:

Use the free fall equation to find the acceleration due to gravity For an object dropped from a height to find velocity

Answer 1

The Correct Option is C

We are given that a body is dropped freely from a height of 20 m, and it reaches the surface with a velocity of 31.4 m/s. We need to find the length of a simple pendulum that ticks seconds on the planet. Step 1: Use the free fall equation to find the acceleration due to gravity For an object dropped from a height, the final velocity \(v\) is related to the acceleration due to gravity \(g_{\text{planet}}\) and the height \(h\) by the equation: \[ v^2 = u^2 + 2gh \] where: - \(v = 31.4 \, \text{m/s}\) (final velocity),
- \(u = 0 \, \text{m/s}\) (initial velocity),
- \(h = 20 \, \text{m}\) (height).
Substituting the values: \[ (31.4)^2 = 0 + 2 \cdot g_{\text{planet}} \cdot 20 \] \[ g_{\text{planet}} = \frac{31.4^2}{40} = \frac{985.96}{40} = 24.65 \, \text{m/s}^2 \] Step 2: Relate the acceleration due to gravity to the pendulum's length The period of a simple pendulum is given by: \[ T = 2\pi \sqrt{\frac{L}{g_{\text{planet}}}} \] For the pendulum to tick seconds, the period should be 1 second, so: \[ 1 = 2\pi \sqrt{\frac{L}{24.65}} \] Solving for \(L\): \[ 1 = 2\pi \sqrt{\frac{L}{24.65}} \quad \Rightarrow \quad \frac{1}{2\pi} = \sqrt{\frac{L}{24.65}} \] \[ \left( \frac{1}{2\pi} \right)^2 = \frac{L}{24.65} \] \[ L = \left( \frac{1}{2\pi} \right)^2 \cdot 24.65 \approx 2.525 \, \text{m} \] Thus, the length of the pendulum is \(2.5 \, \text{m}\).