(a) Define gravitational field intensity (b) In an experiment to determine the acceleration of free-fall due to gravity, g, using a simple pendulum of lengt...

Answer Details

(a) Gravitational field intensity at a point is the gravitational force experienced per unit mass placed at that point; \(g = \dfrac{F}{m}\). It is a vector quantity directed towards the centre of the attracting body and is measured in \(\text{N kg}^{-1}\).

(b)(i) Since \(l = \dfrac{g}{4\pi^2}\,T^2\), the graph of \(l\) (vertical axis) against \(T^2\) (horizontal axis) is a straight line passing through the origin with a positive, constant slope:

(b)(ii) For a simple pendulum of length \(l\) oscillating through a small angle, the period is

\[ T = 2\pi \sqrt{\frac{l}{g}} \]

Squaring both sides,

\[ T^2 = \frac{4\pi^2}{g}\, l \quad\Rightarrow\quad l = \frac{g}{4\pi^2}\, T^2 \]

Comparing with \(l = (\text{slope})\,T^2\), the slope of the \(l\)-against-\(T^2\) graph is

\[ \text{slope} = \frac{g}{4\pi^2} \]

(b)(iii) Given slope \(= 0.25\):

\[ g = 4\pi^2 \times \text{slope} = 4 \times (3.142)^2 \times 0.25 \] \[ g = 4 \times 9.872 \times 0.25 = 9.87\ \text{ms}^{-2} \]

(c) Horizontal projection: initial horizontal velocity \(u = 15\ \text{ms}^{-1}\), horizontal range \(R = 45\ \text{m}\), \(g = 10\ \text{ms}^{-2}\).

(c)(i) The horizontal motion is at constant velocity, so the time of flight is

\[ R = u\,t \quad\Rightarrow\quad t = \frac{R}{u} = \frac{45}{15} = 3\ \text{s} \]

(c)(ii) Vertically the stone starts from rest and falls freely, so the height of the wall is

\[ h = \frac{1}{2}\,g\,t^2 = \frac{1}{2} \times 10 \times 3^2 = \frac{1}{2} \times 10 \times 9 = 45\ \text{m} \]

Therefore the time of flight is \(3\ \text{s}\) and the wall is \(45\ \text{m}\) high.