(a) Explain what is meant by acceleration of free fall due to gravity, g. (b) State two reasons why g varies on the surface of the earth (c) A stone is proj...

(a) The acceleration of free fall due to gravity, g, is the constant acceleration produced in a body that is falling freely under the action of gravity alone (no air resistance). It is the rate of increase of velocity of a freely falling body, having an approximate value of \( 9.8\,\text{m s}^{-2} \) (taken as \( 10\,\text{m s}^{-2} \) here).

(b) Two reasons g varies over the earth:

• The earth is not a perfect sphere; its radius is greater at the equator than at the poles, so g is smaller at the equator (since \( g \propto 1/r^2 \)).
• Variation of altitude (height above sea level) - g decreases with increasing distance from the earth's centre.

(c) Resolve the initial velocity: \( u_x = 100\cos 30^\circ = 86.6\,\text{m s}^{-1} \), \( u_y = 100\sin 30^\circ = 50\,\text{m s}^{-1} \).

(i) Maximum height above the ground. Height risen above the tower top: \( H = \dfrac{u_y^2}{2g} = \dfrac{50^2}{2\times 10} = 125\,\text{m} \). Above the ground: \( 125 + 100 = 225\,\text{m} \).

(ii) Time to reach maximum height. \( t = \dfrac{u_y}{g} = \dfrac{50}{10} = 5\,\text{s} \).

(iii) Time of flight. Taking downward displacement to the ground as \(-100\,\text{m}\): \[ -100 = 50t - \tfrac{1}{2}(10)t^2 \] \[ 5t^2 - 50t - 100 = 0 \Rightarrow t^2 - 10t - 20 = 0 \] \[ t = \dfrac{10 + \sqrt{100 + 80}}{2} = \dfrac{10 + 13.42}{2} = 11.7\,\text{s}. \]

(iv) Horizontal distance to Q. \( x = u_x \times t = 86.6 \times 11.7 = 1.01 \times 10^{3}\,\text{m} \) (about 1014 m).